GIFT  OF 


: 


WELLS'S  MATHEMATICAL  SERIES. 


Academic  Arithmetic. 

Essentials  of  Algebra. 

Algebra  for  Secondary  Schools. 

Text-book  in  Algebra. 

Academic  Algebra. 

New  Higher  Algebra. 

Higher  Algebra. 

University  Algebra. 

College  Algebra. 

Advanced  Course  in  Algebra. 

Essentials  of  Plane  Geometry. 

Essentials  of  Solid  Geometry. 

Essentials  of  Plane  and  Solid  Geometry. 

Complete  Trigonometry. 

New  Plane  Trigonometry. 

New  Plane  and  Spherical  Trigonometry. 

Four-Place  Logarithmic  Tables. 

Six-Place  Logarithmic  Tables. 


D.   C.    HEATH   &   CO.,  PUBLISHERS. 


NEW 


PLANE  AND  SPHERICAL 


TRIGONOMETRY 


BY 


WEBSTER  WELLS,  S.B. 

PROFESSOR  OF  MATHEMATICS   IN  THE   MASSACHUSETTS 
INSTITUTE  OF  TECHNOLOGY 


D.    C.   HEATH   &    CO.,   PUBLISHERS 

BOSTON        NEW  YORK        CHICAGO 

1907 


COPYRIGHT,  1896, 
BY  WEBSTER  WELLS. 


PREFACE. 


IN  revising  his  Plane  and  Spherical  Trigonometry,  the  author  has 
effected  many  important  improvements.  The  attention  of  teachers  is 
specially  invited  to  the  following  features  of  the  new  work : 

1.  The  proofs  of  the  functions  of  0°,  90°,  180°,  and  270° ;  §§  22  to  25. 

2.  The  proofs  of  the  functions  of  120°,  135°,  etc. ;  §  27. 

3.  The  method  of  finding  the  values  of  the  remaining  functions  of  an 
angle  when  the  value  of  any  one  is  given ;  §  28. 

4.  The  proofs  of  the  functions  of  (—  A),  and  (90°  +  A),  in  terms  of 
those  of  A;  §§  29,30. 

5.  The  method  of  solution  in  the  examples  of  §§  34  and  35. 

6.  The   general  demonstration  of  the  formulae  for  sin  (x  +  y)   and 
cos  (x  +  y) ;  §  42. 

7.  The  discussion  of  the  line  values  of  the  functions,  and  their  appli- 
cation in  tracing  the  changes  in  the  six  principal  functions  of  an  angle 
as  the  angle  increases  from  0°  to  360°;  §§  60,  61. 

8.  The  discussion  of  trigonometric  equations  in  §  62. 

9.  The  solution  of  right  triangles  by  Natural  Functions ;  see  Ex.  1, 
page  54. 

10.  The  discussion  of  the  ambiguous  case  in  the  solution  of  oblique 
triangles;  §§  117  to  120. 

11.  The  proof  of  the  formulae  for  the  values  of  x  in  the  cubic  equation 
tf  -  ax  -  b  =  0 ;  §  126. 

12.  The  geometrical  proof  of  the  important  theorems  of  §  133. 

13.  The  demonstration  of  the  formulae  for  right  spherical  triangles 
before  those  for  oblique  spherical  triangles ;  see  Chapters  XI.  and  XII. 

14.  The  reduction  of  the  number  of  cases  in  the  complete  demonstra- 
tion of  the  fundamental  theorems  for  spherical  right  triangles,  to  three, 
by  application  of  the  theorems  of  §  133 ;  see  §  136. 

iii 

201944 


iv  PREFACE. 

15.  The  solution  of  Quadrantal   and   Isosceles   Spherical  triangles; 
§§  149,  150. 

16.  The  discussion  of  the  ambiguous  cases  in  the  solution  of  oblique 
spherical  triangles;  §§  165,  166;  especially  the  rules  given  on  pages  108 
and  111  for  determining  the  number  of  solutions. 

At  the  end  of  Chapter  XII.  will  be  found  a  collection  of  formulae  in 
form  for  convenient  reference. 

The  revised  work  contains  a  much  greater  number  of  examples  than 
the  old ;  they  have  been  selected  with  great  care,  and  are  with  few  excep- 
tions new. 

The  results  have  been  worked  out  by  aid  of  the  author's  new  Six  Place 
Logarithmic  Tables,  which  contain  also  a  Table  of  Natural  Functions,  and 
an  Auxiliary  Table  for  Small  Angles.  The  Trigonometry  can  be  obtained 
either  with  or  without  the  Tables. 

WEBSTEK  WELLS. 
MASSACHUSETTS  INSTITUTE  OF  TECHNOLOGY,  1896. 


CONTENTS. 


PLANE  TRIGONOMETRY. 
I.  TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES 

II.  TRIGONOMETRIC  FUNCTIONS  OF  ANGLES  IN  GENERAL    .    .    .    .    /  6 

TIT.   GENERAL  FORMULA /.  19 

IV.  MISCELLANEOUS  THEOREMS        29 

V.  LOGARITHMS       . 41 

Properties  of  Logarithms 43 

Applications 47 

Exponential  Equations 51 

VI.   SOLUTION  OF  RIGHT  TRIANGLES         54 

Formulae  for  the  Area  of  a  Right  Triangle 60 

VII.   GENERAL  PROPERTIES  OF  TRIANGLES 62 

Formulae  for  the  Area  of  an  Oblique  Triangle 65 

VEIL   SOLUTION  OF  OBLIQUE  TRIANGLES 67 

Area  of  an  Oblique  Triangle        74 

IX.  CUBIC  EQUATIONS 77 


SPHERICAL  TRIGONOMETRY. 

X.   GEOMETRICAL  PRINCIPLES 80 

XI.  RIGHT  SPHERICAL  TRIANGLES        83 

Solution  of  Right  Spherical  Triangles       88 

XII.   OBLIQUE  SPHERICAL  TRIANGLES        96 

General  Properties  of  Spherical  Triangles 96 

Napier's  Analogies        101 

Solution  of  Oblique  Spherical  Triangles        102 

Applications Ill 


FORMULAE;  PLANE  TRIGONOMETRY 114 

SPHERICAL  TRIGONOMETRY .  116 

ANSWERS .    , 119 

T 


PLANE    TRIGONOMETRY. 


I.  TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES. 

1.  Trigonometry  treats  of  the  properties  and  measurement  of  angles 
and  triangles. 

In  Plane  Trigonometry  we  consider  plane  figures  only. 

2.  Definitions  of  the  Trigonometric  Functions  of  Acute  Angles. 

Let  BAG  be  any  acute  angle. 

B 


From  any  point  in  either  side,  as  B,  draw  a  perpendicular  to  the  other 
side,  forming  the  right  triangle  ABC. 

We  then  have  the  following  definitions,  applicable  to  either  of  the 
acute  angles  A  or  B : 

In  any  right  triangle, 

The  sine  Of  either  acute  angle  is  the  ratio  of  the  opposite  side  to  the 
hypotenuse. 

The  cosine  is  the  ratio  of  the  adjacent  side  to  the  hypotenuse. 
The  tangent  is  the  ratio  of  the  opposite  side  to  the  adjacent  side. 
The  cotangent  is  the  ratio  of  the  adjacent  side  to  the  opposite  side. 
The  secant  is  the  ratio  of  the  hypotenuse  to  the  adjacent  side. 
The  cosecant  is  the  ratio  of  the  hypotenuse  to  the  opposite  side. 

We  also  have  the  following  definitions : 
The  versed  sine  of  an  angle  is  1  minus  the  cosine  of  the  angle. 
The  coversed  sine  is  1  minus  the  sine. 

The  eight  ratios  defined  above  are  called  the  Trigonometric  Functions 
of  the  angle. 


2  PLANE  TRIGONOMETRY. 

Eepresenting  the  sides  BC,  CA,  and  AB  by  a,  6,  and  c,  respectively, 
and  employing  the  usual  abbreviations,  we  have : 


a 
sin  J.  =  — 

tan^  =  -- 

sec  -4=-- 

vers  J.  =  l 

b 

c 

6 

& 

c 

cos  A  =  — 

cot  -4  =  — 

c 

covers  -4  =  1 

a 

c 

a 

~~a 

c' 

sin£=*. 

b 

sec  B  =  -- 

vers  B  =  l 

a 

C 

~~  a 

a 

G 

cos  B  =  — 

cot  JB  =  -- 

cscB=-- 

covers  B  =  1 

b 

c 

b 

b 

c 

3.  It  is  important  to  observe  that  the  values  of  the  trigonometric 
functions  depend  solely  on  the  magnitude  of  the  angle,  and  are  entirely 
independent  of  the  lengths  of  the  sides  of  the  right  triangle  whiqh  con- 
tains it. 


B/ 


For  let  B  and  B'  be  any  two  points  in  the  side  AD  of  the  angle  DAE, 
and  draw  JBO.and  B'C'  perpendicular  to  AE. 
Then  by  the  definition  of  §  2,  we  have 

BC  B'C' 

sinJ.  =  — -,  and  sinJ.  =  - 

AB?  AB' 

,  But  since  the  right  triangles  ABO  and  AB'C'  are  similar,  their  homol- 
ogous sides  are  proportional ;  whence, 

BC=B'C' 
AB~  AB'' 

Thus  the  two  values  obtained  for  sin  A  are  equal. 

4.   We  have  from  §  2, 

abb  a 

-.  cosJ.  =  -,  smJB  =  -,  and  cos5  =  — 
c  c  <£  c 

Whence,   a  =  c,  sin  JU=  c  cos  B,  and  b  =  c  sin  B  =  c  cos  Al 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES. 


3 


That  is,  in  any  right  triangle,  either  side  about  the  right  angle  is  equal  to 
the  hypotenuse  multiplied  by  the  sine  of  the  opposite  angle,  or  by  the  cosine 
of  the  adjacent  angle. 

Again,         tanJ.  =  -,  cot  A  =  -,  tanJ3  =  -,  and  cotB  =  -- 
baa  b 

Whence,      a  =  6  tan  A  =  6  cot  B,  and  b  =  a  tan  B  =  a  cot  A. 

That  is,  in  any  right  triangle,  either  side  about  the  right  angle  is  equal 
to  the  tangent  of  the  opposite  angle,  or  the  cotangent  of  the  adjacent  angle, 
multiplied  by  the  other  side. 


5.  We  have  from  §  2, 

sin  A  =  -  =  cos  B. 


sec  A  =  -=cscB. 


tan  A  =  -  =  cot  B.  vers  A  =  1  —  =  covers  B. 

b  c 

As  B  is  the  complement  of  A,  these  results  may  be  stated  as  follows : 
The  sine,  tangent,  secant,  and  versed  sine  of  any  acute  angle  are  respec- 
tively the  cosine,  cotangent,  cosecant,  and  coversed  sine  of  the  complement  of 
the  angle. 

6.  To  Find  the  Value?  of  the  Other  Seven  Functions  of  an  Acute  Angle, 
when  the  Value  of  Any  One  is  Given. 

1.    Given  esc  A  =  3 ;  find  the  values  of  the  remaining  functions  of  A* 

B 


3V2 

a 

We  may  write  the  equation  cscN-4  =•-• 

Since  the  cosecant  is  the  hypotenuse  divided  by  the  opposite  side,  we 
may  regard  A  as  one  of  the  acute  angles  of  the  right  triangle  ABC,  in 
which  the  hypotenuse  AB  =  3,  and  the  opposite  side  BG  =• 1. 

Whence  by  Geometry,  AC  =  ^AB*-B(?  =  V9^I  =  V8  =  2  V2. 
Then  by  the  definitions  of  §  2, 


tan^l 


cos  A  =s 


2V2 


1 

2V2* 
cot  A  =  2  V2. 


sec  .4  = 


2V2 


vers  -4  =  1  — 


2V2 


1     2 

covers  J.  =  1  —  -  =  -• 
o      6 


PLANE  TRIGONOMETRY. 


"A- 


2.   Given  vers  A  =  - ;  find  the  value  of  cot  A. 
5 


Since  vers  A  =  1  —  cos  A,  we  have  cos  A  =  1  —  vers  A  =  1  —-  =  -.» 

5     5 

Then,  in  the  right  triangle  ABC,  we  take  the  adjacent  side  AC  =  3, 
and  the  hypotenuse  J_B  =  5. 


Whence,       BC  = 
Then  by  definition, 


25  -  9  =  Vl6  =  4. 


o 

cot  -4  =  — 


EXAMPLES. 
In  each  of  the  following,  find  the  values  of  the  remaining  functions  : 


3. 


5. 


~ 


\|  7.   CosJ=^. 


9. 


4.  vers  A  =  —  •      6.  esc  A  —  I. 
lo 


8.  covers  A=—>       10. 
17 


3  21 

11.  Given  cot  A  —  -;  find  sin  A.          14.   Given  cos  A—  — -;  find  esc  A. 

12.  Given  esc  A  —  — ;  find  cos  A.      V15.   Given  tan  A=— — ;  find  sec  ^ 

40  I 

13.  Given  sec  A  =  5;  find  cot  A.          16.   Given  sin  A=  -;  find  tan  A. 

x 


7.  Functions  of  45°. 


Let  ABC  be  an  isosceles  right  triangle,  AC  and  jB<7  being  each  equal 

to  1.  - 

Then  Z^l  =  450,  and  AB  =  ^AC2  +  BC2  =  VI  4- 1  =  V2. 


TRIGONOMETRIC  FUNCTIONS  OF  ACUTE  ANGLES. 

Whence  by  definition, 

sec  45°  =  V2. 

csc45°=V2. 

vers  45°  =  1  —  —  V2  =  — 
tan  45°  =  1. 


sin  45°  =  —  =  £V2. 

V2     2 

cos45°=-L=fv2. 


8.  Functions,  of  30°  and  60°. 


Let  ^.BD  be  an  equilateral  triangle  having  each  side  equal  to  2. 

Draw  AC  perpendicular  to  BD. 

Then  by  Geometry,  BC  =  £  £Z>  =  1,  and  Z  B^LO  =  ^  Z  J5AD  =  30°. 

Also,  ^(7  =  V^2_^O2=vT^l=V3. 

Then  by  definition,  in  the  right  triangle  ABC, 


sin30°  =  i 
2 


cos  30°  = 


=  cos  60°. 
=  sin  60°. 


sec  30°=— =?V3  =  esc  60°. 
V3    3 


esc  30°=2 


=  sec  60°. 


tan  30°  =  —  =  \  V3  =  cot  60°. 
V3     3 

cot30°  =  V3  =  tan  60°. 


._J^       =  covers  60°o 

2 


covers  30°=l-i=i     =  vers  60°. 

2i     2 


* 


6  PLANE  TRIGONOMETRY. 

II.    TRIGONOMETRIC  FUNCTIONS  OF  ANGLES  IN 

GENERAL. 

9.  In  Geometry,  we  are,  as  a  rule,  concerned  with  angles  which  are 
less  than  two  right  angles;  but  in  Trigonometry  it  is  convenient  to 
consider  them  as  unrestricted  in  magnitude. 


A'" 


Let  AA"  and  A1  A'"  be  a  pair  of  perpendicular  diameters  of  the  circle 
AA". 

Suppose  a  radius  OB  to  start  from  the  position  OA,  and  revolve 
about  the  point  0  as  a  pivot,  in  a  direction  contrary  to  the  motion  of  the 
hands  of  a  clock. 

When  OB  coincides  with  OA ',  it  has  generated  an  angle  of  90°;  when 
it  coincides  with  OA",  of  180°;  with  OA'"9  of  270°;  with  OA,  its  first 
position,  of  360° ;  with  OA  again,  of  450° ;  and  so  on. 

We  thus  see  that  a  significance  may  be  attached  to  a  positive  angle 
of  any  number  of  degrees. 

10.  The  interpretation  of  an  angle  as  measuring  the  amount  of  rota- 
tion of  a  moving  radius,  enables  us  to  distinguish  between  positive  and 
negative  angles. 

Thus,  if  a  positive  angle  indicates  revolution  from  the  position  OA  in 
a  direction  contrary  to  the  motion  of  the  hands  of  a  clock,  a  negative 
angle  may  be  taken  as  indicating  revolution  from  the  position  OA  in  the 
same  direction  as  the  motion  of  the  hands  of  a  clock. 

Thus,  if  the  radius  OB'  starts  from  the  position  OA,  and  revolves 
about  the  point  0  as  a  pivot  in^he  same  direction  as  the  motion  of  the 
hands  of  a  clock,  when  it  coincmes  with'(M"'  it  has  generated  an  angle 
of  -90°;  when  it  coincides  with  OA",  of  -180°;  with  OA',  of  -270°; 
and  so  on. 

We  may  then  conceive  of  negative  angles  of  any  number  of  degrees. 

It  is  immaterial  which  direction  we  consider  the  positive  direction  of 
rotation ;  but  having  at  the  outset  adopted  a  certain  direction  as  positive, 
our  subsequent  operations  must  be  in  accordance. 


TRIGONOMETRIC  FUNCTIONS  OF  ANGLES  IN  GENERAL.        7 

U.  The  fixed  line  OA  from  which  the  rotation  is  supposed  to  com- 
mence, is  called  the  initial  line,  and  the  rotating  radius  in  its  final  position 
is  called  the  terminal  line. 

12.  In  designating  an  angle,  we  shall  always  write  first  the  letter  at 
the  extremity  of  the  initial  line. 

Thus,  in  designating  the  angle  formed  by  the  lines  OA  and  OB,  if 
we  regard  OA  as  the  initial  linef  we  should  call  it  AOB',  and  if  we  regard 
OB  as  the  initial  line,  we  should  call  it  BOA. 

There  are  always  two  angles  less  than  360°  in  absolute  value,  one  posi- 
tive and  the  other  negative,  formed  by  the  same  initial  and  terminal  line. 

Thus,  there  are  formed  by  OA  and  OB1  the  positive  angle  AOB' 
between  270°  and  360°,  and  the  negative  angle  AOB'  between  0°  and 
-90°. 

We  shall  distinguish  between  such  angles  by  referring  to  them  as 
"the  positive  angle  AOB'"  and  "the  negative  angle  AOB',"  respectively. 

13.  It  is  evident  that  the  terminal  lines  of  any  two  angles  which  differ 
by  a  multiple  of  360°  are  coincident. 

Thus,  the  angles  30°,  390°,  -  330°,  etc.,  have  the  same  terminal  line. 

14.  Rectangular  Co-ordinates. 


\  (-b,  a) 

j 

9i  (b,  a) 

a 

a 

, 

b 

ft 

N 

0            M 

a 

a 

P,(-MO 

t 

><(b,-a) 

Y' 

Let  P!  be  any  point  in  the  plane  of  the  lines  XX'  and  YY*  inter- 
secting at  right  angles  at  0,  and  draw  P±M  perpendicular  to  XX'. 

Then  OM  and  P±M  are  called  the  rectangular  co-ordinates  of  Px ;  OM 
is  called  the  abscissa,  and  P±M  the  ordinate. 

The  lines  of  reference,  XX'  and  FT7,  are  called  the  axis  of  X  and  the 
axis  of  Y,  respectively,  and  0  is  called  the  origin. 

It  is  customary  to  express  the  fact  that  the  abscissa  of  a  point  is  &, 
and  its  ordinate  a,  by  saying  that  for  the  point  in  question  x  =  b  and 
y  =  a ;  or,  more  concisely,  we  may  refer  to  the  point  as  "  the  point 
(b,  a),"  where  the  first  term  in  the  parenthesis  is  understood  to  be  the 
abscissa,  and  the  second  term  the  ordinate. 


8 


PLANE  TRIGONOMETRY. 


15.  If,  in  the  figure  of  §  14,  OM  =  ON=  b,  and  PXP4  and  P2P3  are 
drawn  perpendicular  to  XX'  so  that  P^M  —  P2N—  P3N  =  P4M=  a,  the 
points  P1?  P2,  P3,  and  P4  will  have  the  same  co-ordinates,  (b,  a). 

To  avoid  this  ambiguity,  abscissas  measured  to  the  right  of  0  are 
considered  positive,  and  to  the  left,  negative;  and  ordinates  measured  above 
XX'  are  considered  positive,  and  below,  negative. 

Then  the  co-ordinates  of  the  points  will  be  as  follows : 

P,,  (b,  a);  P2,  (-b,  a);  P3,  (-b,  -a);  P4,  (6,  -a). 

16.  If  a  point  lies  upon  XX',  its  ordinate  is  zero ;  and  if  it  lies  upon 
YY,  its  abscissa  is  zero. 

17.  General  Definitions  of  the  Functions. 

We  will  now  give  general  definitions  of  the  trigonometric  functions, 
applicable  to  any  angle  whatever. 

Take  the  initial  line  of  the  angle  as  the  positive  direction  of  the  axis 
of  X,  the  vertex  being  the  origin. 

From  any  point  in  the  terminal  line,  drop  a  perpendicular  to  the  axis 
ofX. 

Find  the  co-ordinates  of  this  point ;  then, 

The  sine  of  the  angle  is  the  ratio  of  the  ordinate  of  the  point  to  its 
distance  from  the  origin. 

The  cosine        is  the  ratio  of  the  abscissa  to  the  distance. 

The  tangent      is  the  ratio  of  the  ordinate  to  the  abscissa. 

The  cotangent  is  the  ratio  of  the  abscissa  to  the  ordinate. 

The  secant        is  the  ratio  of  the  distance  to  the  abscissa. 

The  cosecant     is  the  ratio  of  the  distance  to  the  ordinate. 

18.  We  will  now  apply  the  definitions  of  §  17  to  finding  the  functions 
of  the  angles  XOP2,  XOPS,  and  XOP*  in  the  following  figures : 


2  (-6,0) 


M 


Y' 

Let  P2,  P3,  and  P4  be  any  points  on  the  terminal  lines  OP2,  OP8,  and 
OP4,  and  draw  P2  Jf,  P3M,  and  P4 M  perpendicular  to  XX1. 

Let  P2M  =  P3^T=  P4^f  =  a,  OM=  b,  and  OP2  =  OP8  =  OP4  =  c. 


TRIGONOMETRIC   FUNCTIONS  OF  ANGLES  IN  GENERAL.        9 

Then  the  co-ordinates  of  P2  are  (—6,  a);   of  P3,  (—6,  —a);   of  P^ 
(b,  -a). 

Whence  by  definition, 

sin  XOP,  =  -•  sin  XOP3  =  —  =  -  -•    sin  XOP4  =  —  =  -  -• 

c  c  c  c  c 

cos  XOP2  =  —  =  --•    cos  XOP3  =  —  =  --•    cosXOP4  =  -- 
c  c  c  c  c 

tanXOP2  =  —  =  --•    tanXOP3  =  —  =  -•       tan  XOP4  =  —  =  -  ?• 
—  66  —66  66 

cotXOP2  =  —  =  --•    cotXOP3  =  —  =  -•        cotXOP4  =  —  =  --• 
a  a  —  a     a  —  a         a 


—00  —00 

csc  XOP2  =  -•  csc  XOP3  =  —  =  -  -•    esc  XOP4  =  —  =  -  -• 

a  —a         a  —a         a 

Note  1.     The  definitions  of  §  17  are  seen  to  include  those  of  §  2. 

The  definitions  of  the  versed  sine  and  coversed  sine,  given  in  §  2,  are  sufficiently 
general  to  apply  to  any  angle  whatever. 

Note  2.  In  all  the  figures  of  the  present  chapter,  the  small  letters  will  be  under- 
stood as  denoting  the  lengths  of  the  lines  to  which  they  are  attached,  without  regard 
to  tlidr  algebraic  sign. 

19,  If  the  initial  line  of  an  angle  coincides  with  OX,  and  its  terminal 
line  lies  between  OX  and  OY,  the  angle  is  said  to  be  in  the^rsZ  quadrant; 
if  the  terminal  line  lies  between  0  Y  and  OX',  the  angle  is  said  to  be  in 
the  second  quadrant;  if  between  OX1  and  OY7,  in  the  third  quadrant;  if 
between  0  Y1  and  OX,  in  the  fourth  quadrant. 

Thus,  any  positive  angle  between  0°  and  90°,  or  360°  and  450°,  or  any 
negative  angle  between  —  270°  and  —  360°,  is  in  the  first  quadrant  ;  any 
positive  angle  between  90°  and  180°,  or  450°  and  540°,  or  any  negative 
angle  between  —  180°  and  —  270°,  is  in  the  second  quadrant. 

20.  It  follows  from  the  definitions  of  §  17  that,  for  any  angle  in  the 
first  quadrant,  all  the  functions  are  positive. 

It  is  also  evident  by  inspection  of  the  results  of  §  18  that  : 

In  the  second  quadrant,  the  sine  and  cosecant  are  positive,  and  the  cosine, 
tangent,  cotangent,  and  secant  are  negative. 

In  the  third  quadrant,  the  tangent  and  cotangent  are  positive,  and  the  sine, 
cosine,  secant,  and  cosecant  are  negative. 

In  the  fourth  quadrant,  the  cosine  and  secant  are  positive,  and  the  sine, 
tangent,  cotangent,  and  cosecant  are  negative. 


10  PLANE  TRIGONOMETRY. 

It  is  usual  to  express  the  above  in  tabular  form,  as  follows : 


Functions. 

First 
Quad. 

Second 
Quad. 

Third 
Quad. 

Fourth 
Quad. 

4- 

+ 

Cosine  and  secant                .     •     • 

4. 

4- 

Tangent  and  cotangent    .... 

+ 

— 

+ 

21.  Since  the  terminal  lines  of  any  two  angles  which  differ  by  a  mul- 
tiple of  360°  are  coincident  (§  13),  it  is  evident  that  the  trigonometric 
functions  of  two  such  angles  are  identical. 

Thus,  the  functions  of  50°,  410°,  770°,  -  310°,  etc.,  are  identical. 

22.  Functions  of  0°  and  360°. 


The  terminal  line  of  0°  coincides  with  the  initial  line  OX. 
Let  P  be  a  point  on  OX  such  that  OP  =  a. 
Then  by  §  16,  the  co-ordinates  of  P  are  (a,  0). 
Whence  by  definition, 


sin  0°  =    =  0. 

a 

cosO°  =  -  =  l. 


tanO°  =  - 


0. 


cot  0°  =    = 


secO°  =  2 


cscO°=? 


a  v  0 

By  §  21,  the  functions  of  360°  are  the  same  as  those  of  0°. 

23.  Functions  of  90°. 

T 

P(0,a) 


Let  P  be  a  point  on  OF  such  that  OP=»  a, 
Then  the  co-ordinates  of  P  are  (0,  a). 


TRIGONOMETRIC  FUNCTIONS  OF  ANGLES  IN  GENERAL. 

Whence  by  definition, 


sin  90° 


cos  90°  =  -  =  0. 
a 


tan  90°  =-  =  oo. 

4  ~ 

cot  90°  =  -  =  0. 
a 


sec  90°  =    = 


esc  90°  =  -  =  1. 
a 


24.  Functions  of  180°. 


.180° 


Let  P  be  a  point  on  OX'  such  that  OP  =  a. 
Then  the  co-ordinates  of  P  are  (—  a,  0). 
Whence  by  definition, 


sin  180°  =-  =  0. 
a 

cos  180°  =  :=^  =  -l. 
a 


tan  180°  = 
cot  180°  = 


0 


—  a 

—  a 


=  00. 


sec  180°  = 
esc  180°  = 


—  a 


=  -1. 


25.  Functions  of  270°. 


270° 


P(0,-a) 
Y' 

Let  P  be  a  point  on  OT  such  that  OP=  a. 
Then  the  co-ordinates  of  P  are  (0,  —  a). 
Whence  by  definition, 

sin  270°  =  —  =  - 1.         tan  270°  =  —  =  oo.          sec  270°  =  -  =  oo. 
a  0    -  0 


cos  270°  =    =  0. 
a 


cot  270°  =  —  =  0. 
—  a 


esc  270°  =        =  _i. 
—  a 


Note.    No  absolute  meaning  can  be  attached  to  such  a  result  as  cotO°  =  oo  ;  it 
merely  signifies  that  as  an  angle  approaches  0°,  its  cotangent  increases  without  limit. 
A  similar  interpretation  must  be  given  to  the  equations  esc  0°=  oo,  tan  90°=  oo,  etc. 


12 


PLANE   TRIGONOMETRY. 


26.   The  results  of  the  last  four  articles  may  be  conveniently  expressed 
in  tabular  form  as  follows : 


Angle. 

Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

0° 

0 

1 

0 

00 

1 

00 

90° 

1 

0 

00 

0 

00 

1 

180° 

0 

-1 

0 

00 

-1 

00 

270° 

-1 

0 

00 

0 

00 

-1 

360° 

0 

1 

0 

00 

1 

00 

27.  Functions  of  120°,  135°,  150°,  etc. 


Let  0PM  be  a  right  triangle  having  OP,  OM,  and  PM  equal  to  2, 1, 
and  V3,  respectively,  and  Z  POM=  60°.     (Compare  §  8.) 

Then  Z  XOP=  120°,  and  the  co-ordinates  of  P  are  (-  1,  V3). 
Whence  by  definition, 

sin  120°  =  —.         tanl20°  =  -V3.  sec  120°  =  -2. 


cosl20°  =  -. 


cotl20°  =  -  —  =  - 
V3         3 


csc  120°  =  -^:  =  |  V3. 
V3     3 


In  like  manner  may  be  proved  the  remaining  values  given  in  the  fol- 
lowing table,  which  are  left  as  exercises  for  the  student  : 


.AwgrZe. 

Sin. 

Cos. 

Tan. 

Co«. 

Sec. 

Csc. 

120° 

kVS 

-4 

-V3 

-1  V3 

-2 

|V3 

135° 

-JV2 

^ 

-l 

-1 

-V2 

V2 

150° 

1 

-|-\/3 

-iV3 

-V3 

-|V3 

2 

210° 

—  1 

—  |V3 

iV3 

V3 

-|V3 

-2 

225° 

—  iV2 

—  iV2 

1 

1 

-V2 

-V2 

240° 

—  £V3 

"2" 

V3 

|V3 

-2 

-|VS 

300° 

-|V3 

1 

-V3 

—  JV3 

2 

—  |V3 

315° 

—  iV2 

-1 

-1 

V2 

-V2 

330° 

-* 

iV3 

-iV3 

-V3 

|V3 

-2 

TRIGONOMETRIC  FUNCTIONS  OF  ANGLES  IN  GENERAL.     13 


28.  Given  the  value  of  one  function  of  an  angle,  to  find  the  values  of  the 
remaining  functions.  (Compare  §  6.) 

1.   Given  sin  A  =  —  -  ;  find  the  values  of  the  remaining  functions  of  A. 
5 

The  example  may  be  solved  by  a  method  similar  to  that  of  §  6 ;  since 
the  sine  is  the  ratio  of  the  ordinate  to  the  distance,  we  may  regard  the 
point  of  reference  as  having  its  ordinate  equal  to  —  3,  and  its  distance 
equal  to  5. 

There  are  two  points,  P  and  P',  which  are  3  units  below  the  axis  of 
X,  and  distant  5  units  from  0. 


There  are  then  two  angles,  XOP  and  XOP1,  in  the  third  and  fourth 
quadrants,  respectively,  either  of  which  may  be  the  angle  A. 

Now,  0 M  =  O M*  =  V  OP2  _  PM* 


=  V25-9  =  4. 

Then  the  co-ordinates  of  P  are  (—  4,  —  3) ;  and  of  P',  (4,  —  3). 
Whence  by  definition : 


Angle. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

XOP 

4 

3 

4 

5 

5 

5 

4 

3 

4 

3 

XOP' 

4 
5 

3 
4 

4 
3 

5 
4 

5 
3 

Thus  the  two  solutions  to  the  problem  are: 


cos  .4  =  T      tan 


±3,         u 


=  T-,  esc  J.  =  --; 


where  the  upper  signs  refer  to  XOP,  and  the  lower  signs  to  XOP' 

2.   Given  cot  A  =  3;  find  the  values  of  the  remaining  functions  of  A. 

o  o 

The  equation  may  be  written  either  cot  A  ==  -,  or  cot  A  =  —  *?• 


14 


PLANE  TRIGONOMETRY. 


We  may  then  regard  the  point  of  reference  as  having  its  abscissa 
equal  to  3  and  its  ordinate  equal  to  1,  or  as  having  its  abscissa  equal  to 
—  3  and  its  ordinate  equal  to  —  1. 

There  are  two  angles,  XOP  and  XOP',  in  the  first  and  third  quad- 
rants, respectively,  either  of  which  satisfies  the  given  condition. 


P'C-8,-1) 


Then  OP=  OP' =  ^OM2  +  PM2  =  V9+T  =  VlO. 
Whence  by  definition : 


Angle. 

Sin. 

Cos. 

Tan. 

Sec. 

Csc. 

YY)P 

1 

3 

1 

Vio 

l-\  (\ 

"EDP' 

Vio 
i 

Vio 

3 

3 
1 

3 

Vio 

vio 

/TA 

Vio 

Vio 

3 

3 

—  VIO 

Thus  the  two  solutions  are  : 


,  cosA=± 


=,  sec  A= 


)  esc  A=±  VlO. 


Vio7  'Vio7 

Note.     It  must  be  clearly  borne  in  mind,  in  examples  like  the  above,  that  the 
"distance"  is  always  positive. 

EXAMPLES. 

In  each  of  the  f ollowing,  find  the  values  of  the  remaining  functions : 

3.  sec^.  =  ^.  7.    csc^  =  -—  -  11.   tan -4  =  -7. 

4.  cot  A  =  —  — •  8.   tan  A  =  — •  12-    esc  A  =  3. 


21 
29' 


9. 
10. 


13.  cos^  =  -- 

o 

14.  cot  A  =  x. 


TRIGONOMETRIC  FUNCTIONS  OF  ANGLES  IN   GENERAL.     15 
29.  Functions  of  (—A)  in  terms  of  those  of  A. 


M 
P' 


FIG.  1. 


FIG. 


There  may  be  four  cases :  A  in  the  first  quadrant  (Fig.  1),  A  in  the 
second  quadrant  (Fig.  2);  A  in  the  third  quadrant  (Fig.  3),  or  A  in  the 
fourth  quadrant  (Fig.  4). 

In  each  figure,  let  the  positive  angle  XOP  represent  the  angle  A,  and 
the  negative  angle  XOP'  the  angle  —  A. 

Draw  PM  perpendicular  to  XX1 ,  and  produce  it  to  meet  OP'  at  P'. 

In  the  right  triangles  0PM  and  OP'M,  the  side  OM  is  common,  and 
Z  POM=  Z  P'OM. 

Hence,  the  triangles  are  equal,  and  PM=  P'M  and  OP  =  OP'. 

Then  in  each  figure, 

abscissa  P'  =      abscissa  P, 

ordinate  P'  =  —  ordinate  P, 

and  distance  P'  =      distance  P. 


Then, 
ord.  P' 

ord.  P 

ord.  P' 

ord.P 

dist.  P' 

dist.  P 

dist.P' 
abs.  P' 

dist.  P 
abs.P 

abs.  P' 
abs.  P' 

abs.P 
abs.P 

abs.  P' 
dist.  P' 

abs.P 
dist.  P 

dist.  P' 

dist.P 

ord.  P' 

ord.P 

ord.  P' 

ord.P 

Whence, 

sin  (—  A)  =  —  sin  A. 
cos(—A)='    cos  ^4.. 


tan  (—  A)  =  —  tan  A. 
cot  (—A)  =  —  cot  A. 


sec(—  A)=     sec  A.  | 
esc  (—  -4)  =  —  esc  A. ) 


... 


16  PLANE  TRIGONOMETRY. 

30.  Functions  of  (90°  -f  A)  in  terms  of  those  of  A. 
P'       Y  P 

90  °v 


M' 


P' 


FIG.  1. 


FIG.  2. 


FIG. 


There  may  be  four  cases :  A  in  the  first  quadrant  (Fig.  1),  A  in  the 
second  quadrant  (Fig.  2),  A  in  the  third  quadrant  (Fig.  3),  or  A  in  the 
fourth  quadrant  (Fig.  4). 

In  each  figure,  let  the  positive  angle  XOP  represent  the  angle  A>  and 
the  positive  angle  XOP'  the  angle  90°  +  A. 

Take  OP'  =  OP,  and  draw  PM  and  P'M'  perpendicular  to  XX'. 

Since  OP  is  perpendicular  to  OP',  and  OM to  PM' ,  Z  POM=Z  0PM1. 

Then  the  right  triangles  0PM  and  OP'M'  have  the  hypotenuse  and  an 
acute  angle  of  one  equal  to  the  hypotenuse  and  an  acute  angle  of  the  other. 

Hence,  the  triangles  are  equal,  and  PM=  OM1  and  OM=  P'M9. 

Then  in  each  figure, 

ordinate  P'  =     abscissa  P, 

abscissa  P'  =  —  ordinate  P, 

and  distance  P'  =      distance  P. 


Then, 


ord.  P' 
dist.  P' 
abs.  P' 
dist.  P' 
ord.  P' 
abs.  P' 


3.  P 


abs.  P'         ord.  P 


dist.  P 
ord.  P 


ord.  P' 
dist.  P' 
abs.  P 
dist.  P' 
ord.  P' 


abs.  P 
dist.  P 
ord.  P* 
dist.  P 


TRIGONOMETRIC   FUNCTIONS  OF  ANGLES  IN  GENERAL.     17 

Or,   sin  (90°  +  A)  =     'cos  A.        cot  (90°  +  A)  =  -  tan  A.  •) 

cos  (90°  +  A)=  —  sin  A.        sec  (90°  +  A)=  —  esc  A.  I  (2) 

tan  (90°  +  A)  =  -  cot  A.        esc  (90°  +  A)  =     sec  A  J 

31.  The  results  of  §  30  may  be  stated  as  follows  : 

The  sine,  cosine,  tangent)  cotangent,  secant,  and  cosecant  of  anyjzngle  are 
equal,  respectively,  to  the  cosine,  minus  the  sine,  minus  the  cotangent,  minus 
the  tangent,  minus  the  cosecant,  and  the  secant,  of  an  angle  90°  less. 

32.  Functions  of  (90°  -  A)  in  terms  of  those  of  A. 

By  §  31,   sin  (90°  -  A)  =      cos  (-  A)  =  cos  A  (§  29). 

cos  (90°  -  A)  =  -  sin  (-  A)  =  sin  A. 

tan  (90°  —  A)=—  cot  (-  A)  =  cot  A. 

cot  (90°  -  A)  =  -  tan  (-  A)  =  tan  A. 

sec  (90°  —  A)=—csc(—A)=  esc  A. 

esc  (90°  —  A)=      SQC(-A)  =  sec  A. 
These  formulae  were  proved  for  acute  angles  in  §  5. 

33.  Functions  of  (180°  -  A)  in  terms  of  those  of  A. 

By  §  31,  sin  (180°  -  A)  =  cos  (90°  -A)=  sin  A  (§  32). 
cos  (180°  -A)=-  sin  (90°  -  A)  =  -  cos  A. 
tan  (180°  -  A)  =  -  cot  (90°  -  ^1)  =  -  tan  A. 
cot  (180°  -  ^4)  =  -  tan  (90°  —  A)=—  cot  A. 
sec  (180°  -  A)  =  -  esc  (90°  -A)=-  sec  A 
esc  (180°  -A)=  sec  (90°  -  ^)  =  esc  A. 

34.  By  successive  applications  of  the  theorem  of  §  31,  any  function  of 
a  multiple  of  90°,  plus  or  minus  A,  may  be  expressed  as  a  function  of  A. 

1.  Express  sin  (270°  +  A)  as  a  function  of  A. 

By  §  31,  sin  (270°  +  A)  =  cos  (180°  +  A)  =  -  sin  (90°  +  A)=-  cos  A. 
If  the  multiple  of  90°  is  greater  than  270°,  we  may  subtract  360°,  or 
any  multiple  of  360°,  from  the  angle,  in  accordance  with  §  21. 

2.  Express  sec  (990°  —  A)  as  a  function  of  A. 
Subtracting  twice  360°,  or  720°,  from  the  angle,  we  have 

sec  (990°  -A)  =  sec  (270°  -  A). 
And  by  §  31,        sec  (270°  -A)=-  esc  (180°  -  A)  =  -  esc  A  (§  33). 

If  the  multiple  of  90°  is  negative,  we  may  add  360°,  or  any  multiple 
of  360°,  to  the  angle. 


18  PLANE  TRIGONOMETRY. 

3.  Express  tan  (— 180°  +  A)  as  a  function  of  A. 
Adding  360°  to  the  angle,  we  have 

tan  (-  180°  +  A)  =  tan  (180°  +  -4). 
And  by  §  31,       .   tan  (180°  +  A)  =  -  cot  (90°  +  A)  =  tan  A.    ' 

EXAMPLES. 
Express  each  of  the  following  as  a  function  of  A : 

4.  sin  (180° +  4).  9.   sec  (630°  +  A).  14.  tan (-450° -4). 

5.  cos  (270°  -A).         10.  tan  (-270° --4).  15.  cos  (-  900°  -  A). 

6.  cot  (450°  +  A).         11.   csc  (-90° -A).  16.  sin  (810°  -4). 

7.  osc(360° -4).         12.   cot  (-180°  +  A).  17.  esc  (1080°  +  4). 

8.  tan (540°  -4).         13.   sin  (-  630°  +  4).  18.  sec  (1260°  +  4). 

35.  By  means  of  the  theorem  of  §  31,  any  function  of  any  angle,  posi- 
tive or  negative,  may  be  expressed  as  a  function  of  a  certain  acute  angle. 

1.  Express  sin  317°  as  a  function  of  an  acute  angle. 
By  §  31,  sin  317°  =  cos  227°  =  -  sin  137°  =  -  cos  47°. 

Since  the  complement  of  47°  is  43°,  another  form  of  the  result  is 
-sin 43°  (§5). 

Note.    As  in  the  examples  of  §  34,  360°,  or  any  multiple  of  360°,  may  be  added 
to,  or  subtracted  from,  the  angle. 

EXAMPLES. 

Express  each  of  the  following  as  a  function  of  an  acute  angle : 

2.  cos  322°.  4.   sec  559°.  6.   cot  (-378°). 

3.  tan 208°.  5.   esc  803°  45'.  7.   sin(-139°5'). 

It  is  evident  from  the  above  that  any  function  of  any  angle  can  be 
expressed  as  a  function  of  a  certain  acute  angle  less  than  45°. 

Express  each  of  the  following  as  a  function  of  an  acute  angle  less  than 

45°: 

8.  cot  155°.  10.   sec  457°.  12.  tan  (-681°). 

9.  sin  1138°  36'.  11.   cos  496° 20'.  13.   esc  (-257°). 

14.  Find  the  value  of  esc  (-  210°). 
Adding  360°  to  the  angle,  we  have 

esc  (-210°)=  esc  150°. 
And  by  §  31,  esc  150°  =  sec  60°  =  2  (§  8). 

Find  the  values  of  the  following : 

15.  cot 405°.  17.  esc  600°.  19.  cos  (-420°). 

16.  sin  480°.  18.  tan  690°.  20.   sec  (-225°). 


GENERAL  FORMULA. 


19 


III.    GENERAL  FORMULA. 

36.  It  follows  immediately  from  the  definitions  of  §  17  that,  if  x  is 
any  angle, 


cosa;  = 


37. 


1 

fort  /r 

qpp  rp  — 

csc  a; 
1 

cot  a; 
1 

cos  a; 

pqp  7*  — 

sec  a; 
e  the  formula 

tana/ 

sm  a? 

tana; 


sma; 
cos  a; 


I.  When  the  angle  x  is  acute  (or  in  the  first  quadrant). 


Fia.  1. 


In  the  right  triangle  ABC,  let  BAG  be  the  angle  x. 

BO 


By  §2, 


BC     AB     sin  a; 

tan  x  =  -— —  =  — — •  = 


AC     AC     cos  x 

AB 
II.  When  x  is  in  the  second,  third,  or  fourth  quadrant. 


M 


-X       * 


r     x'- 


y 


Fie.  2. 


FIG.  8. 


In  each  figure,  let  the  positive  angle  XOP  represent  the  angle  a?,  and 
draw  PM  perpendicular  to  XX'. 

Then  in  each  figure,  by  the  definitions  of  §  17, 

ord.  P 

t        =  ord'  P=  dist<  P  =  sin  x 
abs.  P     abs.  P     cos  a?' 

dist.  P 


20  PLANE  TRIGONOMETRY. 

38.    To  prove  the  formula 


cote,  =        .  (5) 

sm  x 


By  (3),  §  36,  cota;  =  -  -  =  -L  (§  37)  = 

tan  x     sma  v        '      sm  a; 

COSOJ 

39.    To  prove  the  formula 

sin2  a;  -f  cos2  a?  =  1.  (6) 

Note.     Sin2  x  signifies  (sin  sc)2  j  that  is,  the  square  of  the  sine  of  «. 

I.   When  the  angle  x  is  acute  (or  in  the  first  quadrant). 
In  Fig.  1,  §  37,  we  have  by  Geometry, 


Dividingby^,  f 


Then  by  definition,       (sin  x)2  +  (cos  x)2  =  1. 
That  is,  sin2  x  +  cos2  x  =  1. 

II.   When  a;  is  in  the  second,  third,  or  fourth  quadrant. 
In  each  of  the  figures  2,  3,  and  4  of  §  37,  we  have 


PM2  +  OM2  =  OP2. 


Dividing  by  OP2, 


OP2      OP2 

2  2 


But  in  either  figure,  2  =  sin2  x,  and  •  -  -  =  cos2  x. 

Whence,  sin2  x  -\-  cos2  2=1. 

Formula  (6)  may  be  written  in  the  forms 

sin2  x  =  1  —  cos2  x,  and  cos2  x  =  1  —  sin2  a?. 

40.    To  prove  the  formulae 

sec2a;  =  l  +  tan2a;,  (7) 

2nd                                       esc2  x  =  1  +  cot2  a;.  (8) 

By  (6),                                     1  =  cos2  x  +  sin2  a?.  (A) 

Dividing  by  cos2  x,  -V  =  *  +  ?^' 

cos2  a;  cos2  a; 

Whence  by  (3)  and  (4),  sec2  a*  =  1  -f  tan2  a?. 


GENERAL  FORMULAE, 


21 


Again,  dividing  (A)  by  sin2  x,  we  have 

1     __  i  i  cos2  a?. 
sin2  a? 


snic 


Whence  by  (3)  and  (5),  esc2  x  =  1  -f  cot2  x. 

41.    To  find  the  values  of  sin  (x  +  y)  and  cos  (x  +  y)  in  terms  of  the  sines 
find  cosines  of  x  and  y. 

y 

I.  When  x  and  y  are  acute,  and  x  -f  y  acute. 

c/ 


0  ul     D 

Let  ^40.5  and  JBO(7  denote  the  angles  x  and  y,  respectively. 

Then,  Z  .400  =x  +  y. 

From  any  point  (7  in  OC  draw  (7J.  and  CB  perpendicular  to  OA  and 
OB ;  and  draw  .BZ>  and  BE  perpendicular  to  OA  and  AC. 

Since  -EJO  and  BC  are  perpendicular  to  <M  and  OB,  the  angles 
and  AOB^TQ  equal ;  that  is,  Z  BCE  =  #. 


Now, 


But, 


and 


and 


Whence,       sin  (x  -\-  y)  —  sin  x  cos  y  +  cos  x  sin  y. 
Again,          cos  ( 

But, 


00 


_ 
00 


_ 
00     00 


Whence,       cos  (a;  +  y)  =  cos  #  cos  y  —  sin  a;  sin  y. 


(9) 


(10) 


22  PLANE  TRIGONOMETRY. 

II.  When  x  and  y  are  acute,  and  x  +  y  obtuse. 

V 


0    J> 


Let  DOB  and  BOG  denote  the  angles  x  and  y,  respectively. 

Then,  Z.DOC=x  +  y. 

From  any  point  C  in  OC  draw  CB  perpendicular  to  OB,  and  CA 
perpendicular  to  DO  produced;  and  draw  BD  and  BE  perpendicular 
to  OD  and  AC. 

Since  EC  and  BC  are  perpendicular  to  OZ)  and  OB,  the  angles 
and  DOB  are  equal ;  that  is,  Z.  BCE  ==  x. 


But, 


and 


Whence, 
Again, 


ft3»j$~"*» 

sin  (x  +  y)  =  sin  a;  cos  y  +  cos  a;  sin  y. 


cos  DOC  = 


-OA     OD-  BE     OD     BE 


OC 


OC 


OC     OC 


But, 


§§=§*§— 


and 


Whence,       cos  (x  +  y)  =  cos  x  cos  y  —  sin  x  sin  y. 

42.  Formulae  (9)  and  (10)  are  very  important,  and  it  is  necessary  to 
prove  them  for  all  values  of  x  and  y. 

They  have  already  been  proved  when  x  and  y  are  any  two  acute 
angles ;  or,  what  is  the  same  thing,  when  they  are  any  two  angles  in  the 
first  quadrant. 

Now  let  a  and  b  be  any  assigned  values  of  x  and  y,  for  which  (9)  and 
(10)  are  true. 


GENERAL  FORMULA.  23 

By  (2),  §  30,       sin  [90°  +  (a  +  6)]  =     cos  (a  +  6), 
and  cos  [90°  +  (a  +  &)]=-  sin  (a  +  6). 

Whence,  by  (9)  and  (10), 

sin  [90°  -f  (a  +  &)]  =      cos  a  cos  b  —  sin  a  sin  6,  (A) 

and  cos  [90°  +  (a  +  6)]  =  —  sin  a  cos  b  —  cos  a  sin  6.  (B) 

But  by  (2),  §  30,  cos  a  =  sin  (90°  +  a),  and  —  sin  a  =  cos  (90°  +  a). 
Then,  (A)  and  (B)  may  be  written  in  the  forms 

sin  [(90°  +  a)  +  6]  =  sin  (90°  -f  a)  cos  b  +  cos  (90°  +  a)  sin  6, 
and      cos  [(90°  +  a)  +  &]  =  cos  (90°  +  a)  cos  b  -  sin  (90°  +  a)  sin  b  ; 

which  are  in  accordance  with  (9)  and  (10). 

Therefore,  if  (9)  and  (10)  hold  for  any  assigned  values  of  *  and  y, 
they  also  hold  when  one  of  the  angles  is  increased  by  90°. 

But  they  have  been  proved  to  hold  when  x  and  y  are  both  in  the 
first  quadrant;  hence,  they  hold  when  x  is  in  the  second  quadrant  and 
y  in  the  first. 

And  since  they  hold  when  x  is  in  the  second  quadrant  and  y  in  the 
fiT,of  +-u~~  T_I  j  -  -|ien  x  an^  y  are  ^0^  ^  the  second  quadrant;  and  so  on. 

Hence,  (9)  and  (10)  hold  for  any  values  of  x  and  y  whatever,  positive 
or  negative. 

43.   By  (9),  sin  [x  +,(—  #)]  =  sin  ^  cos  (~  #).+  cos  »  sin  (—  y) 

—  sin  x  cos  y  +  cos  «(—  sin  y),  by  (l),  §  29. 
(a?  —  y)  —  sin  a?  cos  y  —'cos  x  sin  y.  (11) 

By  (10),        cos  [x  -f  (—  y)~]  =  cos  g  cos  (—  y)  —  sin  a;  sin  (—  2/) 

=  cos  a;  cos  y  —  sin  #  (—  sin  ?/). 
cos  (#  —  y)  =  jos  a;  cos  $r  +  sin  &  sin  ?/.  (12) 


^  sin  s  cos  y  +  COSE  sin  y   .      (g) 
cos  a?  cos  y  —  sin  a;  sin  y 


Dividing  each  term  of  the  fraction  by  cos  x  cos  y, 

sin  x  cos  y  ,  cos  a;  sin  y 
cos  x  cos  y     cos  x  cos  i/ 


tan  (x  4-  y)  = 

'      cos  x  cos  y  _  sin  x  sin  ?/ 

cos  x  cos  y     cos  x  cos  ?/ 
tan  a;  4-  tan  y 
1  —  tan  x  tan  y 


24  PLANE  TRIGONOMETRY. 

In  like  manner,  we  may  prove 


Again,by(5), 


__  cos  x  cos  y  —  sin  x  sin  y 
sin  x  cos  y  +  c°s  *  sin  y 
Dividing  each  term  of  the  fraction  by  sin  x  sin  y, 

cos  x  cos  y     sin  a;  sin  y 


sin  a;  cosy,  cos  a;  sin  y 
sin  x  sin  y      sin  x  sin  y 
_.  cotftcoty-1 
cot  y  4-  cot  x 

In  like  manner,  we  may  prove 


sn  s  sn  y     sn  g  sin  y 

,lg. 

(16) 

cot  y  —  cot  a 

45.  From  (9),  (10),  (11),  and  (12),4  we  have 

"  SID  (a  -f-  &)  =  sin  a  cos  &  4-  cos  a  sr  (A) 

v.  sin  (a  —  6)  =  sin  a  cos  6  —  cos  (B) 

f  cos  (a  4-  &)  =  cos  a  cos  b  —  sin  a  sin  6. 
.cos  (a  —  b)  =  cos  a  cos  6  4-  sin  a  si;  (B) 

Adding  and  subtracting  (A)  and  (B),  and  then  (C)  and  (D), 
t 

sin  (a  4  &)  +  sin  (a  —  6)  =     2  sin  a  cos  b. 

sin  (a  -f  &)  —  sin(a  —  6)  =  2  cos  a  sin  b. 
cos(a-f  &)  +  cos(a  —  5)=  2cosacos&. 
cos  (a  -f  6)  —  cos  (a  —  b)  =  —  2  sin  a  sin  6. 

Let  a  -f  6  =  »,  and  a  —  b  =  y. 

Then,  a  =  £(»  +  $,  and  6  =  ^(a?-2/). 

Substituting  these  values,  we  have 

sina-f  siny=     2  sin  %  (x  +  y)  cos  $(x  —  y).  (17) 

sin  aj  —  sin  y  =     2  cos  -J-  (a;  4-  y)  sin  £  (cc  —  y).  (18) 

cos  a;  4-  cosy  =     2  cos  -J-  (a;  4-  y)  cos  £  (#  —  y).  (19) 

cos  a  —  cosy  =  —  2  sin  £  (a;  4-  y)  sin  £  (a?  —  y).  (20) 


GENERAL  FORMULA.  25 


46.  By  (17)  and  (18),  we  have 

sin  x  +  sin  y  _  2  sin  %  (x  +  y)  cos  j-  (a;  —  y) 
sin  a?  —  sin  y  ~~  2  cos  %  (x  +  y)  sin  ^  (a?  —  y) 

a;  —  y) 


47.  By  (9)  and  (11),  we  have 

sin  (a?  +  y)  sin  (x  —  y)=  (sin  a;  cos  #  -j-  cos  x  sin  y)  (sin  x  cos  y  —  cos  x  sin  y) 
=  sin2  x  cos2*  y  —  cos2  x  sin2  y 
=  sin2  x  (1  —  sin2  y)  —  (1  —  sin2  #)  sin2  1/  (§  39) 
=  sin2  x  —  sin2  x  sin2y  —  sin2  y  4-  sin2  #  sin2  y 
=  sin2  a;  —  sin2  y.  (22) 

The  result  may  also  be  written 

sin  (a;  -f-  y)  sin  (a;  —  y)  =  1  —  cos2  x  —  (1  —  cos2  y)  (§  39) 

=  cos2  y  —  cos2  x.  (23) 

In  like  manner,  we  may  prove 

cos  (x  -f  y)  cos  (x  —  y)=  cos2  cc  —  sin2  y  =  cos2  y  —  sin2  aj.  (24) 

48.  Functions  of  2  x. 

Putting  y  =  a;  in  (9),  we  have 

sin  2  x  =  sin  a;  cos  a?  +  cos  a;  sinoj 

=  2  sin  a;  cos  a;.  (25) 

Putting  y  =  x  in  (10),  we  obtain  ( 

cos  2  a;  =  cos2  a?  —  sin2  a;.  (26) 

We  also  have  by  §  39, 

cos  2  x  =  (1  -  sin2  x)  -  sin2  x  =  1  -  2  sin2a;,  (27) 

and  cos  2  x  —  cos2  a;  —  (1  —  cos2  x)  —  2  cos2  a?  —  1.  (28) 

Putting  y  =  a;  in  (13)  and  (15),  we  have 


49.  Functions  of  Jr. 

From  (27)  and  (W)  we     LVC,  by  transposition, 
os2a?,  and  2cos2a;  = 


26  PLANE  TRIGONOMETRY. 

Putting  %x  in  place  of  x,  and  therefore  x  in  place  of  2#,'we  have 

2  sin2^  x  =  1  —  cos  x,  (31) 

2  cos2  £  a;  =  1  +  cos  x.  (32) 

Again,  putting  ^x  in  place  of  a;  in  (25), 

2  sin  £  x  cos  £  a;  =  sin  x.  (A) 

Dividing  (31)  by  (A),  we  have,  by  (4), 


smx 


(33) 


Dividing  (32)  by  (A),      cot£a;  =  i±  !2§£.  (34) 

50.  Functions  of  3x. 

We  have,  sin  3a;  =  sin  (2x  +  x)=  sin2a?  cos  x  +  cos2a;  sina;,  by  (9) 
=  (2  sin  x  cos  a;)  cos  x -f (1 — 2  sin2  a?)  sin  4^§  48) 
=  2  sina; (1  —  sin2a;)+  sina;  —  2  sin3a;^39) 
=3  sina;  —  4  sin3 x.  (35) 

Also,         cos  3  x  =  cos  (2  x  -f  a;)  =  cos  2  a?  cos  a;  —  sin  2  a;  sin  x}  by  (10) 
=  (2  cos2  a;— l)cosa;—  (2  sin  a;  cos  a;)  sin  a;  (§  48) 
=  2  cos3 a?  —  cos  x  —  2  cosa;(l  —  cos2 a;)  (§  39) 
=  4  cos3  a;  —  3  cos  x. 


Again,      ^3-^(8.  +  .)-  by  (X3) 

2tana; 


—  tan2a;  ,      x 

-,  by  (29) 


-fcs 


tan  x\ 


tana; 


—  2  tan  a;  +  (1  —  tan2  a?)  tan  a;  _  3  tan  x  —  tan3  a; 
l-tan2a;-2tan2a;  l-3tan2a?   ' 

EXERCISES. 
51.  1.  Prove  the  relation  sec2  a;  esc2  a;  =  sec2  a;  +  esc2  a;. 


By  (3),  sec^  **»  =      ...    =  ,  by  (6) 

cos2  x  sin2  x        cos2  x  siir  x 

sin2  a?  cos2  a; 


cos"  a;  sin"  a;     cos"  x  siir  a; 


cos  x     sin  a; 


GENERAL   FORMULAE.  27 


2.  Prove  the  relation  sin3a?~  sina?  =  tana. 
cos  3  a;  +  cosa? 


.    sin  3  x  —  sin  x    2  cos  \  (3  x  -f  x)  sin  J  (3  x  —  x) 
By  (18)  and  (19) 


3.   Prove  the  relation     tan  (a  +  y)  -  tan  a;  = 
1  -f-  tan  (x  +  y)  tan  a; 


Prove  the  following  relations  : 

4    sin  (a;  +  y)  _  tan  a;  -f  tan  y  ~    cos  (a;  4-  y)  _  1  —  tan  x  tan  y 

sin  (x  —  y)~  tan  a?  —  tan  y  '   cos  (x  —  y)      1  -f-  tan  a;  tan  i/ 

=  _co      (  )ooti(a,_    }. 


cos  x  —  cosy 

7.  sin  (oj  +  y  H-  2;)  =  sin  a;  cos  y  cos  z  4-  cos  a;  sin  y  cos  2 

+  cos  x  cos  y  sin  2;  —  sin  x  sin  ?/  sin  z. 

8.  cos  (a?  4-  y  +  2)  =  cos  x  cos  ?/  cos  z  —  sin  a;  sin  y  cos  2 

—  sin  x  cos  2/  sin  z  —  cos  x  sin  2/  sin  z. 

9.  tan(60°+a;)-cot(300-a;)=0.       12     /tan  a?  +'1V=  1  +  sin  2  a; 

\tan  x  —  \)      1  —  sin  2  a; 

10.    -       -  -  -—sec  2  A.  1Q     sin  5  x  +  sin  x  «„ 

esc2  A  —  2  l«i.    -  !  -  =  tan  o  a?. 

cos  5  a;  +  cos  a; 


11.  _=SeCa;.  14.  -  sia  5  g  =  -  cot  4  a; 

sin  x        cos  a;  cos  3  x  —  cos  5  x 

15.  sin  4  a;  =  4  sin  a;  cos  x  —  8  sin3  x  cos  a?. 

16.  cos  4  a;  =  1  —  8  cos2  x  +  8  cos4  x. 

17.  By  putting  x  =  45°  and  y  =  30°  in  (11)  and  (12),  prove 

sin  15°  =  \  (  V6  -  V2),  cos  15°  =  £  (  V6  +  V2). 

18.  By  putting  a;  =  30°  in  (33)  and  (34),  prove 

tan  15°  =  2  -  V3,  cot  15°  =  2  +  V3. 

19.  Using  the  results  of  Ex.  17,  prove 

sec  15°  =  V6  -  V2,  esc  15°  =  V6  -f  V2. 

20.  By  putting  x  =  45°  in  (31)  and  (32),  prove 

sin  221°  =  1.V2-V2,  cos  22^°  = 


28  PLANE  TRIGONOMETRY. 

21.  By  putting  x  =  45°  in  (33)  and  (34),  prove 

tan  22£°  =  V2  -  1,  cot  22  £  =  V2  +  1. 

22.  By  putting   x  —  22y   in  (7)  and  (8),  and  using  the  result  of 
21,  prove 

sec22J°=V4-2V2,  esc  22  £°  =  V4  +  2  V2. 

Prove  the  following  relations  : 

23.  tan  (46°'  -f  a?)  -  tan  (45°  -  a;)  =  2  tan  2  a;. 

24.  cos4  a?  —  sin4  x  =  cos  2  a?.  25.  -  -  -  =  cot  4  a;. 

esc  a?  —  cot  a; 

26.   sin2  (a?  +  y)  —  sin2  (a;  —  y)  =  sin  2  x  sin  2  y. 

27.  tan  (x  -  y)  +  tan  y  .  _  tan  g 

1  —  tan  (a;  —  y)  tan  ?/ 

28.  cos  5  A  cos  3  -4  +  sin  5  A  sin  3  A  =  cos  2  A. 

29.  sin  (  J.  +  B)  cos  (-4  -  £)  -  cos  (A  +  5)  sin  (  A  -  JB)  =  sin  2  5. 

30.  £2^  +  !in^=2cot2a;.  33>    sin  4  a:  +  sin  3  a:  =  CQ 

sin  a;       cos  x  cos  «3  a?  —  cos  4  a; 

31.  sin2a?=    2tan^..  34.   sin  50°  +  sin  10°  =  sin  70°. 

tan2a; 


32.  cos2*  =  .  35.  0    =tana?. 

1  +  tan2  a;  1  -f-  cos  x  +  cos  2  a; 

36.  2  cos  3  a;  sin  x  =  sin  4  a?  —  sin  2  a;. 

37.  cos  5  x  =  5  cos  x  —  20  cos3  x  +  16  cos5  #. 

cos  a;         cota?  +  l  ton  1  a;        4  tan  g  ~ 


1  —  sin  x  ~~  cot^-a;  —  l  1  —  6  tan2  x  -f  tan4  a; 

40.    (sin  x  -f  cos  a;)  (2  —  sin  2  a;)  =  2  (sin3  x  -f  cos3  a?). 


41.    (sin  x  —  sin  y)2+  (cos  a;  —  cos  y)2  =  4  sin 


x  ~ 


42.  -=tan4a;.      43. 


+  sin  a;  +  cos  a;  sina;  +  cos  a; 


MISCELLANEOUS  THEOREMS. 


IV.    MISCELLANEOUS  THEOREMS. 

52.  Circular  Measure  of  an  Angle. 

An  angle  is  measured  by  finding  its  ratio  to  another  angle,  adopted 
arbitrarily  as  the  unit  of  measure. 

The  usual  unit  of  measure  for  angles  is  the  degree,  which  is  an  angle 
equal  to  the  ninetieth  part  of  a  right  angle. 

Another  method  of  measuring  angles,  and  one  of  great  importance,  is 
known  as  the  Circular  Method;  in  which  the  unit  of  measure  is  the  angle 
at  the  centre  of  a  circle  subtended  by  an  arc  whose  length  is  equal  to  the 
radius. 


Thus,  let  AOB  be  any  angle;  and  let  AOC  be  the  unit  of  circular 
measure;  that  is,  the  angle  at  the  centre  subtended  by  an  arc  whose 
length  is  equal  to  OA. 

Then,  circular  measure  AOB  = 


But  by  Geometry, 

Z.AOG     arc  AC        OA 

Whence,  circular  measure  AOB     ar 


That  is,  the  circular  measure  of  an  angle  is  the  ratio  of  its  subtending 
arc  to  the  radius  of  the  circle. 

53.  By  §  52,  the  circular  measure  of  a  right  angle  is  the  ratio  of  one- 
fourth  the  circumference  to  the  radius. 

But  if  R  denotes  the  radius,  the  circumference  of  the  circle  is  2  irR. 

Whence,         circular  measure  of  90°  ==  iof  2?rR  =  1L. 

R  2 

It  follows  from  the  above  that  the  circular  measure  of  180°  is  IT  ;  of 

60°,  |  ;  of  45°,  |  ;  etc. 

That  is,  an  angle  expressed  in  degrees  may  be  reduced  to  circular  meas- 
ure by  finding  its  ratio  to  180°,  and  multiplying  the  result  by  IT. 


Thus,  since  115°  is  |f  of  180°,  the  circular  measure  of  115°  is 

OD  OO 


30  PLANE  TRIGONOMETRY. 

54.  Conversely,  an  angle  expressed  in  circular  measure  may  be  reduced 
to  degrees  by  multiplying  by  180°  and  dividing  by  TT  ;  or,  more  briefly,  by 
substituting  180°  for  IT. 

Thus,  l^^Lof  180?  =  84°. 
15      15 


55.  In  the  circular  method,  such  expressions  may  occur  as  "  the  angle 
-§,"  "the  angle  1,"  etc. 

These  refer  to  the  unit  of  circular  measure ;  thus,  the  angle  -|  signifies 
an  angle  whose  subtending  arc  is  two-thirds  of  the  radius. 

The  angle  1,  that  is,  the  angle  whose  subtending  arc  is  equal  to  the 
radius,  or  the  unit  of  circular  measure,  reduced  to  degrees  by  the  rule  of 
§  54,  gives 

*  =  57.2958°,  approximately. 


7T 

Then  the  rule  of  §  54  may  be  modified  as  follows : 
An  angle  expressed  in  circular  measure  may  be  reduced  to  degrees  by 
multiplying  by  57.2958°. 

Thus,  the  angle  |  =  |  x  57.2958°  =  38.1972°  =  38°  11'  49.92". 

EXAMPLES. 
56.   Express  each  of  the  following  in  circular  measure : 

1.  120°.  3,   67°  30'.  5.   86°  24'.  7.   163°  7' 30". 

2.  315°.  4.   146°  15'.  6.   53°  20'.  8.   88°  53' 20". 

Express  each  of  the  following  in  degree  measure : 

9.  — .  11.   — •  13.  -.  15.   7r~1. 

6  64  4  6 


l°-1f         12-|  14-i- 

57.  Inverse  Trigonometric  Functions. 

The  expression  sin"1^,  called  the  inverse  sine  of  #,  or  the  anti-sine  of 
a?,  signifies  the  angle  whose  sine  is  x. 

Thus,  the  statement  that  the  sine  of  the  angle  x  is  equal  to  y  may  be 
expressed  in  either  of  the  ways 

sin  x  =  y\  or  x  =  sin-1  y. 

In  like  manner,  cos"1  x  signifies  the  angle  whose  cosine  is  x ;  tan"1  #, 
the  angle  whose  tangent  is  x ;  etc. 

Note.  The  student  must  be  careful  not  to  confuse  the  above  notation  with  the 
exponent  —  1 ;  the  —  1  power  of  sinx  is  expressed  (sinx)"1,  and  not  sin^z. 


MISCELLANEOUS  THEOREMS. 


31 


It  is  evident  that  the  sine  of  the  angle  whose  sine  is  x  is  x ;  that  is, 
sin  (sin"1  x)  =  x. 

In  like  manner,  cos  (cos"1  x)  —  x ;  tan  (tan"1  x)  =  x ;  etc. 

58.   By  aid  of  the  principles  of  §  57,  we  may  derive  from  any  formula 
involving  direct  functions  a  relation  between  inverse  functions. 

1.   Prom  the  formula  tan  (x  +  y)  =  tana?  +  tany  ,  prove 

1  —  tan  x  tan  y 

tan-1  a  +  tan'1  b  =  tan'1  J°L±A. 
1  —  ab 

Let  tan  x  =  a,  and  tan  y  =  6. 

Then  by  §  57,  x  =  tan"1  a,  and  y  =  tan"1 6. 

Substituting  these  values  in  the  given  formula, 


tan  (tan-1  a  +  tan-1  b)  = 


Whence, 


a  +  b 
1-ab 

tan-1  a  -f-  tan-1 6  =  tan-1  ^Lt — 
1  —  ab 


2.  Prove  the  relation     cot'1  a  -  sec-1  b  =  cos'1    ^^- -> 

b  Va2  4- 1 
Let  cot"1  a  =  a?,  and  sec"1 6  =  y. 


Then, 
Now, 


cot  x  ==  a,  and  sec  y  =  b. 
cos  (x  —  y)  =  cos  a;  cos  y  -h  sin  a;  sin  ?/. 


(A) 


To  find  the  values  of  the  sines  and  cosines  of  x  and  y,  we  may  use  the 
method  of  §  6. 


In  the  right  triangle  containing  the  angle  a?,  the  adjacent  side  is  a, 
and  the  opposite  side  1 ;  then,  the  hypotenuse  is  Va2  +  1. 

In  the  right  triangle  containing  the  angle  y,  the  hypotenuse  is  6,  and 
the  adjacent  side  1 ;  then,  the  opposite  side  is  V&2  —  1. 

Substituting  the  values  of  cos  x,  cos  y,  sin  x,  and  sin  y  in  (A),  we  have 

cos  (a;  -  y)  = 
Whence,    x  —  yoi  cot-1  a  —  sec"1 6  =  cos"1  — 


32 


PLANE  TRIGONOMETRY. 


EXAMPLES. 


cot2  oc  —  1 

3.   From  the  formula  cot  2  x  =  — ,  prove 

2  cot  a; 


2  cot"1  a  =  cot 


2a 


4.  From  the  formula  cos  2  #  =  1  —  2  sin2  a?,  prove 

2  sin-1  a  =  cos-1  (1-2  a2). 

5.  From  the  formula  sin  2  a?  =  2T^in  x  cos  a?,  prove 

2  cos-1  a  =  sin-1  (2  aVl  -  a2). 

6.  From  the  formula  cos  (x  +  y)  =  cos  a;  cos  y  —  sin  a  sin  y,  prove 

cos"1  a  +  cos-1  b  =  cos"1  (a&  —  VI  —  a2  Vl  —  fc2). 

7.  From  the  formula  sin  3  x  =  3  sin  x  —  4  sin3  x,  prove 

3  sin-1  a  =  sin-1  (3  a  —  4  a3). 

Prove  the  following  relations  : 

8.  cot-1  a  +  cot-1  b  =  cot-ia6  ~  *• 


9.  2  cos-1  a  =  cos-1  (2  a2  -  1). 

10.  sin-1  a  —  sin-1  6  =  sin-1  (aVl  —  &3  —  6  VI  —  a2). 

11.  Stan-a-tan-f^g. 

12.  cot-1  (a  -  6)-  cot-1  (a  +  V)=  cot-lffl2~ 


±i  0 


13.    rin^a  +  cos-1    = 


14.  sec"1  a  —  esc"1  b  =  coe" 

15.  tan"1  a  +  cos"1-  =  sin 
16. 

17.  : 
18. 


-- 
a  —  1 


ab 

-il-^-iQ+Va'-l. 
a  Va2  + 1 

=  taa'12v 


1  —  z  a 


—  o  —  &  ao 


MISCELLANEOUS  THEOREMS. 


33 


59.  The  following  table  expresses  the  value  of  each  of  the  six  prin- 
cipal functions  of  an  angle  in  terms  of  the  other  five : 


sin  A 
cos  A 
tan  A 
cot  A 
sec  A 
esc  A 


sin  .4 


cos  A 
cos  A 


1 


cos  A 
1 


tan  A 


tan  A 


tan  ^4. 


sec  A 

1 
sec  A 


1 
esc  A 


— 1 


The  reciprocal  forms  were  proved  in  §  36. 

The  others  may  be  derived  by  aid  of  §§  36,  37,  38,  39,  and  40,  and  are 
left  as  exercises  for  the  student. 

As  an  illustration,  we  will  give  a  proof  of  the  formula 


cos  A  = 


VcscM  —  1 

—  --       .         II  .  M.I 

csc^l 


By  §  39, 


csc  A 


They  may  also  be  conveniently  proved  by  the  method  of  §  6 ;  thus,  let 
it  be  required  to  prove  the  formula  for  each  of  the  other  functions  in 
terms  of  the  secant. 

We  have 


Since  the  secant  is  the  ratio  of  the  hypotenuse  to  the  adjacent  side,  we 
take  AB  =  sec  A,  and  AC  =  1  j  whence,  BO  =  ^/AS-A&=  Vsec2^  - 1. 


34 


PLANE  TRIGONOMETRY. 


Then  by  definition, 

sin  A  =         c ^— . 

sec^L      ' 


tan  A  =  VsecM  —  1, 
1 


VsecM  —  1 
60.  Line  Values  of  the  Functions. 


esc  A  — 


sec^l 


Let  AGE  be  any  angle.  With  0  as  a  centre,  and  a  radius  equal  to  1, 
describe  the  circle  AB\  draw  BD  and  AE  perpendicular  to  XX',  and  CF 
perpendicular  to  YY1.  Then  by  §  17,  the  functions  of  AOB  are : 


Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

Fi2  1 

BD 

OD 

BD 

OD 

OB 

OB 

J.  lg.  _L. 

"Rio-  9 

OB 
BD 

OB 
OD 

OD 
BD 

BD 
OD 

OD 
OB 

BD 
OB 

JD  lg.  £l. 

TTio-  3 

OB 
BD 

OB 
OD 

OD 
BD 

BD 
OD 

OD 
OB 

BD 
OB 

-C  Ig.  O. 

Pio-  4. 

OB 
BD 

OB 
OD 

OD 
BD 

BD 
OD 

OD 
OB 

BD 
OB 

J?lg.  1. 

OB 

OB 

OD 

BD 

OD 

BD 

MISCELLANEOUS  THEOREMS. 


35 


But  since  the  right  triangles  OBD,  OEA,  and  OOF  are  similar,  and 
OA=  00=1,  we  have 


==Q 
OD      OA 


OD      OA 


=        =  C 
BD     OC 


BD      OC 

Whence,  since  OB  =  1,  the  functions  of  AOB  are : 


Sin. 

Cos. 

Tan. 

Cot. 

Sec. 

Csc. 

Fig.  1. 

BD 

OD 

AE 

OF 

OE 

OF 

rig.  2. 

BD 

-OD 

-AE 

-OF 

-OE 

OF 

Fig.  3. 

-BD 

-OD 

AE 

OF 

-OE 

-OF 

Fig.  4. 

-BD 

OD 

-AE 

-CF 

OE 

-OF 

That  is,  if  the  radius  of  the  circle  is  1, 

The  sine  is  the  perpendicular  drawn  to  XX'  from  the  intersection  of 
the  circle  with  the  terminal  line. 

The  cosine  is  the  line  drawn  from  the  centre  to  the  foot  of  the  sine. 

The  tangent  is  that  portion  of  the  geometrical  tangent  to  the  circle  at 
its  intersection  with  OX  included  between  OX  and  the  terminal  line,  pro- 
duced if  necessary. 

The  cotangent  is  that  portion  of  the  geometrical  tangent  to  the  circle 
at  its  intersection  with  0  T  included  between  0  Y  and  the  terminal  line, 
produced  if  necessary. 

The  secant  is  that  portion  of  the  terminal  line,  or  terminal  line  pro- 
duced, included  between  the  centre  and  the  tangent. 

The  cosecant  is  that  portion  of  the  terminal  line,  or  terminal  line  pro- 
duced, included  between  the  centre  and  the  cotangent. 

And  with  regard  to  algebraic  signs, 

Sines  and  tangents  measured  above  XX'  are  positive,  and  below,  nega- 
tive; cosines  and  cotangents  measured  to  the  right  of  YY'  are  positive, 
and  to  the  left,  negative;  secants  and  cosecants  measured  on  the  terminal 
line  itself  are  positive,  and  on  the  terminal  line  produced,  negative. 

The  above  are  called  the  line  values  of  the  trigonometric  functions. 

They  simply  represent  the  values  of  the  functions  when  the  radius  is 
1 ;  that  is,  the  numerical  value  of  the  sine  of  an  angle  is  the  same  as  the 
number  which  expresses  the  length  of  the  perpendicular  drawn  to  XX' 
from  the  intersection  of  the  circle  and  terminal  line. 


36 


PLANE  TRIGONOMETRY. 


61.    To  trace  the  changes  in  the  six  principal  trigonometric  functions  of 
an  angle  as  the  angle  increases  from  0°  to  360°. 


Let  the  terminal  line  start  from  the  position  OA,  and  revolve  about 
the  point  0  as  a  pivot,  in  a  direction  contrary  to  the  motion  of  the  hands 
of  a  clock. 

Then  since  the  sine  of  the  angle  commences  with  the  value  0,  and 
assumes  in  succession  the  values  -BiA?  B*D&  OC,  B3D8)  J54D4,  etc.  (§  60), 
it  is  evident  that,  as  the  angle  increases  from  0°  to  90°,  the  sine  increases 
from  0  to  1 ;  from  90°  to  180°,  it  decreases  from  1  to  0 ;  from  180°  to  270°, 
it  decreases  (algebraically)  from  0  to  —  1 ;  and  from  270°  to  360°,  it  in- 
creases from  —  1  to  0. 

Since  the  cosine  commences  with  the  value  OA,  and  assumes  in  suc- 
cession the  values  ODlt  OD2)  0,  —  ODS,  —  OD4,  etc.,  from  0°  to  90°/it 
decreases  from  1  to  0 ;  from  90°  to  180°,  it  decreases  from  0  to  —  1 ;  from 
180°  to  270°,  it  increases  from  -1  to  0;  and  from  270°  to  360°,  it  increases 
from  0  to  1. 

Since  the  tangent  commences  with  the  value  0,  and  assumes  in  suc- 
cession the  values  AE»  AE2,  oo,  —  AEB,  —  AE4,  etc.,  from  0°  to  90°,  it 
increases  from  0  to  oo ;  from  90°  to  180°,  it  increases  from  —  oo  to  0 ;  from 
180°  to  270°,  it  increases  from  0  to  oo ;  and  from  270°  to  360°,  it  increases 
from  —  oo  to  0. 

Since  the  cotangent  commences  at  oo,  and  assumes  in  succession  the 
values  CFj,  CF2,  0,  -  CF3)  -  CF4,  etc.,  from  0°  to  90°,  it  decreases  from 
oo  to  0 ;  from  90°  to  180°,  it  decreases  from  0  to  -  oo ;  from  180°  to  270°, 
it  decreases  from  oo  to  0 ;  and  from  270°  to  360°,  it  decreases  from  0  to 

—  00. 

Since  the  secant  commences  with  the  value  OA,  and  assumes  in  suc- 
cession the  values  OE1}  OE2,  oo,  —  OES,  —  OE+,  etc.,  from  0°  to  90°,  it 
increases  from  1  to  oo ;  from  90°  to  180°,  it  increases  from  —  oo  to  —  1 ; 
from  180°  to  270°,  it  decreases  from  - 1  to  -  oo ;  and  from  270°  to  360°, 
it  decreases  from  oo  to  1. 


MISCELLANEOUS  THEOREMS.  37 

Since  the  cosecant  commences  at  oo,  and  assumes  in  succession  the 
values  OFl}  OF2,  OC,  OF3,  OF4)  etc.,  from  0°  to  90°,  it  decreases  from  oo 
to  1  ;  from  90°  to  180°,  it  increases  from  1  to  oo  ;  from  180°  to  270°,  it 
increases  from  —  oo  to  —  1  ;  and  from  270°  to  360°,  it  decreases  from  —  1 
to  —oo. 

Note.  Wherever  the  symbol  co  occurs  in  the  above  discussion,  it  must  be  inter- 
preted as  explained  in  the  Note  to  §  25. 

62.  Trigonometric  Equations. 
1.   Find  the  value  of  A  when  cos  A  =  J. 
We  know  that  one  value  of  A  is  60°  (§  8). 

And  since  cos  (-  60°)=  cos  60°  (§  29),  another  value  of  A  is  —  60°. 
Again,  by  the  principle  of  §  21,  any  multiple  of  360°  may  be  added  to, 
or  subtracted  from,  an  angle,  without  altering  its  functions. 
Hence,  other  values  of  A  are 

360°+60°,  720°+60°,  -360°+60°,  360°-60°,  720°-60°,  -360°-60°,  etc. 

It  is  evident  from,  the  above  that  the  number  of  possible  values  of  A 
is  indefinitely  great  ;  and  that  each  is  in  the  form 

n  x  360°  +  60°,  or  n  X  360°  -  60°  ; 

where  n  is  0,  or  any  positive  or  negative  integer. 
Using  the  circular  notation,  we  have 


o 

2.  Find  the  value  of  A  when  tan  A  =  £  V3. 

We  know  that  one  value  of  A  is  30°  (§  8)  ;  another  is  180°  +30°  (§  27). 
Adding  to,  and  subtracting  from,  these  angles  multiples  of  360°,  other 
values  of  A  are 

360°  +  30°,  540°  +  30°,  -360°  +  30°,  -180°  +  30°,  etc. 

It  is  evident  from  the  above  that  all  the  values  of  A  are  given  by  the 

expression 

nx!80°  +  30°; 

where  n  is  0,  or  any  positive  or  negative  integer. 
Or,  A 


o 

3.  Find  the  value  of  A  when  sin  A  =  -J-  V2. 
One  value  of  A  is  45°  (§  7). 

And  since   sin  (180°  -  45°)  =  sin  45°  (§  33),  another  value  of  A  is 
180°  -  45°. 


38  PLANE  TRIGONOMETRY. 

Adding  to,  and  subtracting  from,  these  angles  multiples  of  360°,  other 
values  of  A  are 

360°  +  45°,  540°  -  45°,  -  360°  +  45°,  -  180°  -  45°,  etc. 

It  is  evident  from  the  above  that  all  the  values  of  A  are  given  by  the 

expression 

nx  180°  +  (-!)"  45°; 

where  n  is  0,  or  any  positive  or  negative  integer. 
Or,  ^  =  mr  +  (-l)»|- 

It  is  evident  that,  to  find  the  value  of  A  in  any  equation  of  the  above 
forms,  we  find  any  one  of  the  values  of  A,  and  substitute  it  for  A  in  the 
following  expressions  : 

If  sin  A  is  given,  nir  +  (—  l)n-4. 
If  cos  A  is  given,  2  mr  ±  A. 
If  tan  A  is  given,  n-n-  +  A. 

The  rule  for  equations  giving  the  value  of  cot  A  is  the  same  as  for 
tan  A  ;  for  sec  A,  the  same  as  for  cos  A  ;  and  for  esc  A,  the  same  as  for 
sin  A. 

EXAMPLES. 

In  each  of  the  following  find  the  value  of  A  : 

4.  tan^l=V3.          6.  sin  A=±.          8.  cot^=-l.          10.  cot  A=0. 

2 

5.  cos  A=  -  JV5.     7.  sec  A=  V2.       9.  esc  A=  -f  VS.    11.  sec  A=  -1. 


63.  1.   Solve  the  equation  cos  2  A  =  cos  A. 

By  (28),     2  cos2  A  —  1  =  cos  A}  or  2  cos2  A  —  cos  A  =  1. 

Solving  this  equation,  cos  A  —     ^        "*  -  =     ^     =1  or  —  -• 

If  cos  .4=1,  one  value  of  A  is  0°  (§  22),  and  A=2  mr  (§  62). 

If  cos  A  =  -  1,  one  value  of  u4  is  120°  (§  27),  and  A  =  2  WTT  ±  ~ 

2.   Solve  the  equation  tan  2x  —  6  tan  a. 


One  solution  is  evidently  tan  x  =  0. 

In  this  case,  one  value  of  x  is  0°,  and  x  =  nir  (§  62). 


MISCELLANEOUS  THEOREMS. 


39 


Dividing  (A)  by  2  tan  x,  we  have 


1  —  tan2  x 


=  3,  or  1  =  3-3  tan2  x,  or  3  tan2  x  =  2. 


2  /2 

Whence,  tan2o;  =  -,  or  tan  x  =  ±  A/-  =  ±  -J  V6. 

* 

Therefore, 


o  *  o 

x  =  tan"1  (  ±  J  V6)  =  ±  tan"1  (£  V6). 


EXAMPLES. 

In  each  of  the  following  find  the  value  of  x : 


3.  sin  x  =  sin  2  x. 

4.  sin  2  x  +  cos  #  =  0. 

5.  cos  x  +  cos  3  x  =  0. 

6.  tan  3  x  -f  tan  x  =  0. 

64.  Limiting  Values  of 


and 


7.  cot  2  x  +  cot  x  =  0. 

8.  tan(45°-a)+cot(45°-a;)=4. 

9.  tan  3  x  =  5  tan  x. 
10.   cos  x  cot  &==!. 
tana; 


To  find  the  limiting  values  of  the  fractions  -  and  -  when  x  is 
indefinitely  decreased. 

Note.     We  suppose  x  to  be  expressed  in  circular  measure  (§  62,  . 

P 


Let  OPXP'  be  a  sector  of  a  circle. 

Draw  PT  and  P'T  tangent  to  the  arc  at  P  and  P',  and  join  OT  and 
PP'. 

By  Geometry,  PT=  P'T. 

Then  OT  is  perpendicular  to  PP'  at  its  middle  point  M,  and  bisects 
the  arc  PP'  at  X. 

Let  Z  XOP  =  Z  XOP'  =  a. 

By  Geometry,    arc  PP'  >  chord  PP',  and  <  PTPf. 

Whence,  arcPX>PJf,  and  <  PT. 


Or  by  §  52,         circ.  meas.  x  >  sin  xt  and  <  tan  x. 


40  PLANE  TRIGONOMETRY. 

Kepresenting  the  circular  measure  of  a;  by  a;  simply,  and  dividing 
through  by  sin  a?,  we  have 

x     .    *         -,     .tancc  1 

>  1.  and  < or  • 

sin  x  sin  x        cos  x 

cni"*  y 

Whence,  —  <  1,  and  >  cos  x. 

MX 

But  when  cc  is  indefinitely  decreased,  cos  a;  approaches  the  limit  1 
(§  22). 

Q1  VJ     rv» 

Hence,  approaches  the  limit  1  when  x  is  indefinitely  decreased. 

tana;       sin  a;       sin  a;        1 

Again,  = = X • 

x        a;  cos  a?        x        cos  a? 

But    and   approach  the  limit  1  when  x  is  indefinitely 

x  cos  a; 

decreased. 

Hence, approaches  the  limit  1  when  x  is  indefinitely  decreased. 


LOGARITHMS.  41 


V.    LOGARITHMS. 

65.  Every  positive  number  may  be  expressed,  exactly  or  approxi- 
mately, as  a  power  of  10. 

Thus,  100  =  102;  13  =  101-113943-  ;  etc. 

When  thus  expressed,  the  corresponding  exponent  is  called  its  Loga- 
rithm to  the  Base  10. 

Thus,  2  is  the  logarithm  of  100  to  the  base  10  ;  a  relation  which  is 
written  Iog10  100  =  2,  or  simply  log  100  =  2. 

66.  Logarithms  of  numbers  to  the  base  10  are  called  Common  Loga- 
rithms, and,  collectively,  form  the  Common  System. 

They  are  the  only  ones  used  for  numerical  computations. 

Any  positive  number,  except  unity,  may  be  taken  as  the  base  of  a 
system  of  logarithms  ;  thus,  if  a*  =  ra,  where  a  and  m  are  positive 
numbers,  then  x  =  loga  m. 

Note.     A  negative  number  is  not  considered  as  having  a  logarithm. 


67.  We  have  by  Algebra, 
10°  =  1, 

10'  =  10, 


10°  =  1,  10-1= 


., 
102  =  100,  10-3  =       =  -001,  etc. 


•• 


Whence  by  the  definition  of  §  65, 

logl  =  0,  log.l  =  -  1  =  9  -  10, 

log  10  =  1,  log .01  =-  2  =  8  -  10, 

log  100  =  2,  log  .001  =  -  3  =?  7  - 10^ etc. 

Note.    The  second  form  for  log.l,  log .01,  etc.,  is  preferable  in  practice.    If  no 
base  is  expressed,  the  base  10  is  understood. 

68.  It  is  evident  from  §  67  that  the  logarithm  of  a  number  greater 
than  1  is  positive,  and  the  logarithm  of  a  number  between  0  and  1 
negative. 

69.  If  a  number  is  not  an  exact  power  of  10,  its  common  logarithm 
can  only  be  expressed  approximately. 

The  integral  part  of  the  logarithm  is  called  the  characteristic,  and  the 
decimal  part  the  mantissa. 


42  PLANE  TRIGONOMETRY. 

.For  example,  log  13  =  1.113943. 

In  this  case,  the  characteristic  is  1,  and  the  mantissa  .113943. 

For  reasons  which  will  appear  hereafter,  only  the  mantissa  of  the 
logarithm  is  given  in  a  table  of  logarithms  of  numbers ;  the  characteristic 
must  be  found  by  aid  of  the  rules  of  §§  70  and  71. 

70.   It  is  evident  from  §  67  that  the  logarithm  of  a  number  between 

1  and      10  is  equal  to  0  +  a  decimal ; 
10  and    100  is  equal  to  1  +  a  decimal ; 
100  and  1000  is  equal  to  2  +  a  decimal ;  etc. 

Therefore,  the  characteristic  of  the  logarithm  of  a  number  with  one 
figure  to  the  left  of  the  decimal  point,  is  0 ;  with  two  figures  to  the  left 
of  the  decimal  point,  is  1 ;  with  three  figures  to  the  left  of  the  decimal 
point,  is  2 ;  etc. 

Hence,  the  characteristic  of  the  logarithm  of  a  number  greater  than  1  is  1 
less  than  the  number  of  places  to  the  left  of  the  decimal  point. 

For  example,  the  characteristic  of  log  906328.51  is  5. 

73-  In  like  manner,  the  logarithm  of  a  number  between 

1  and      .1  is  equal  to  9  +  a  decimal  —  10 ; 
.1  and    .01  is  equal  to  8  +  a  decimal  — 10 ; 
.01  and  .001  is  equal  to  7  +  a  decimal  — 10 ;  etc. 

Therefore,  the  characteristic  of  the  logarithm  of  a  decimal  with  no 
ciphers  between  its  decimal  point  and  first  significant  figure,  is  9,  with 
— 10  after  the  mantissa ;  of  a  decimal  with  one  cipher  between  its  point 
and  first  significant  figure  is  8,  with  —  10  after  the 'mantissa;  of  a  deci- 
mal with  two  ciphers  between  its  point  and  first  significant  figure  is  7, 
with  —  10  after  the  mantissa ;  etc. 

Hence,  to  find  the  characteristic  of  the  logarithm  of  a  number  less  than  1, 
subtract  the  number  of  ciphers  between  the  decimal  point  and  first  significant 
figure  from  9,  writing  —  10  after  the  mantissa. 

For  example,  the  characteristic  of  log  .007023  is  7,  with  —  10  written 
after  the  mantissa. 

Note.  Some  writers  combine  the  two  portions  of  the  characteristic,  and  write 
the  result  as  a  negative  characteristic  before  the  mantissa. 

Thus,  instead  of  7.603658  -  10,  the  student  will  frequently  find  3.603658,  a  minus 
sign  being  written  over  the  characteristic  to  denote  that  it  alone  is  negative,  the 
mantissa  being  always  positive. 


LOGARITHMS. 


PROPERTIES  OF  LOGARITHMS. 

In  any  system,  the  logarithm  of~L  is  0. 
For  by  Algebra,  a°  =  1 ;  whence  by  §  66,  logal  =  0. 

,4L  73.   In  any  system,)  the  logarithm  of  the  base  is  1. 
For  a1  =  a ;  whence,  loga  a  =  1. 

74.  In  any  system  whose  base  is  greater  than  1,  the  logarithm  of  0  is 

—  oo. 

For  if  a  is  greater  than  1,  a~°°  =  —  =  —  =0. 
Whence  by  §  66,  loga  0  =  -  oo  . 

Note.  No  literal  meaning  can  be  attached  to  such  a  result  as  loga  0  =  —  oo  ;  it 
must  be  interpreted  as  follows  : 

If,  in  any  system  whose  base  is  greater  than  unity,  a  number  approaches  the  limit 
0,  its  logarithm  is  negative,  and  increases  without  limit  in  absolute  value. 

75.  In  any  system,  the  logarithm  of  a  product  is  equal  to  the  sum  of  the 
logarithms  of  its  factors. 

Assume  the  equations 

a*  =  m  )  t,     <•/?/?(  ^  = 

[• ;  whence  by  §  66,  4 
a»  =  n  )  '  (y  = 

Multiplying  the  assumed  equations, 

a*  x  a9  =  mn,  or  ax+y  =  mn. 
Whence,  loga  mn  =  x  +  y  =  loga  m  +  loga  n. 

In  like  manner,  the  theorem  may  be  proved  for  the  product  of  three 
or  more  factors. 

76.  By  aid  of  §  75,  the  logarithm  of  a  composite  number  may  be 
found  when  the  logarithms  of  its  factors -are  known. 

1.   Given  log  2  =  .3010  and  log  3  =  .4771 ;  find  log  72. 
Iog72  =  log  (2  x2x2x3x3) 

=  Iog2  +  log 2  4-  Iog2  +  Iog3  +  Iog3  (§  75) 

=  3  x  log  2  +  2  x  log  3  =  .9030  +  .9542  =  1.8572. 


44  PLANE  TRIGONOMETRY. 

EXAMPLES. 

Given  log  2  =  .3010,  log  3  =  .4771,  log  5  ==  .6990,  log  7  =  .8451^  find : 

2.  log  35.  6.   log  126.  10.   log  324.  14.   log  2625. 

3.  log  50.  7.  log  196.  11.   log  378.  15.   log  6048. 

4.  log  42.  8.   log  245.  12.   log  875.  16.   log  12005. 

5.  log  75.  9.   log  210.  13.   log  686.  17.   log  15876. 

77.   In  any  system,  the  logarithm  of  a  fraction  is  equal  to  the  logarithm 
of  the  numerator  minus  the  logarithm  of  the  denominator. 

Assume  the  equations 

>- ;  whence, 
a*  =  n  ) 

Dividing  the  assumed  equations, 


a*      n  n 

Whence,  loga  -  =  x  —  y=  loga  m  —  loga  n. 


78.   1.  Given  log  2  =  .3010;  find  log  5. 

log  5  =  log  —  =  log  10  -  log  2  (§  77)  =  1  -  .3010  =  .6990. 

2 


EXAMPLES. 
Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find : 

2.    log=^-  5.    Iogl4f  8.    log—-  11.    Iog28f. 


3.  log--  6.    log—-  9.    Iog6|.  ,          12.    log^p 

4.  log  45.  7.    log  225.          10.    log  135.  13.    logllOJ. 

' 

79.   In  any  system,  the  logarithm  of  any  power  of  a  quantity  is  equal  to 
the  logarithm  of  the  quantity  multiplied  by  the  exponent  of  the  power. 

Assume  the  equation  ax  —  m ;  whence,  x  =  loga  m. 

R/aising  both  members  of  the  assumed  equation  to  the  pth  power, 

a**  =  mf ;  whence,  logamp  =px=p  logam. 


LOGARITHMS.  45 

i 

80.  In  any  system,  the  logarithm  of  any  root  of  a  quantity  is  equal  to  the 
logarithm  of  the  quantity  divided  by  the  index  of  the  root. 

For,  loga  S/m  =  loga  (m')  =  -  log,  m  (§  79). 

81.  1.    Given  log  2  =  .3010;  find  log  2*. 

log  2*  =  |  x  log  2  =  |  x  .3010  =  .5017. 
o  «j 

Note.    To  multiply  a  logarithm  by  a  fraction,  multiply  first  by  the  numerator,  and 
divide  the  result  by  the  denominator. 

2.    Given  log  3  =  .4771 ;  find  log  -\/3. 


8  8 

EXAMPLES. 
Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find : 

3.  Iog37.  6.   Iog286.  9.   log-\/2.  12.  log  ^525. 

4.  logS*  7.    log  18*.          10.    log  ^6.  13.  log  ^294. 

5.  log  7*.  8.    Iog96l          11.    log  ^7.  14.  Iog^2l6. 

15.  Find  log  (2^x3^). 

By  §  75,  log (2*  X  3*)=  log 2*  +  log 3*  =  •} Iog2  +£ log 3 
=  .1003  +  .5964  =  .6967. 

Find  the  values  of  the  following : 

16.  log  A/I-         18.   log  (2*  x  10*).        20.   log^p.         22. 


V  3  V24 


17.    log(Tj*         19.    log7^X2.  21.   Iog4  23. 

\5/  7? 

82.    To  prove  the  relation 


Assume  the  equations 

ase  =  m)        t  (#  = 

[• ;  whence,  -1 

by  =  m  )  (  v  = 


46  PLANE  TRIGONOMETRY. 

Prom  the  assumed  equations,  a*=&. 

X 

Taking  the  2/th  root  of  both  members,  a?  =  6. 

Therefore,  loga  b  =  -,  or  y  =     x    . 

y  Iog0  b 

That  is,  "* 


83.    To  prove  the  relation 

Iog6  a  x  loga  b  =  1. 
Putting  m  =  a  in  the  result  of  §  82,  we  have 


. 
loga&     loga&  v 

Whence,  Iog6  a  x  loga  6  =  1. 

84.  In  the  common  system,  the  mantissce  of  the  logarithms  of  numbers 
having  the  same  sequence  of  figures  are  equal. 

Suppose,  for  example,  that  log  3.053  =  .484727. 

Then,     log  305.3  =  log  (100  x  3.053)  =  log  100  +  log  3.053 

=  2  +  .484727  =  2.484727  ; 
log  .03053  =  log  (.01  x  3.053)  =  log  .01  +  log  3.053 

=  8  -  10  +  .484727  =  8.484727  -  10  ;  etc. 

It  is  evident  from  the  above  that,  if  a  number  be  multiplied  or  divided 
by  any  integral  power  of  10,  producing  another  number  with  the  same 
sequence  of  figures,  the  mantissas  of  their  logarithms  will  be  equal. 

The  reason  will  now  be  seen  for  the  statement  made  in  §  69,  that  only 
the  mantissse  are  given  in  a  table  of  logarithms  of  numbers. 

For,  to  find  the  logarithm  of  any  number,  we  have  only  to  take  from 
the  table  the  mantissa  corresponding  to  its  sequence  of  figures,  and  the 
characteristic  may  then  be  prefixed  in  accordance  with  the  rules  of  §§  70 
or  71. 

Thus,  if  log  3.053  =  .484727,  then 

log  30.53  =  1.484727,  log  .3053      =  9.484727  -  10, 

log  305.3  =  2.484727,  log  .03053    =  8.484727  -  10, 

log  3053.  =  3.484727,  log  .003053  =  7.484727  -  10,  etc. 

This  property  is  only  enjoyed  by  the  common  system  of  logarithms, 
and  constitutes  its  superiority  over  others  for  the  purposes  of  numerical 
computation. 


LOGARITHMS.  47 

85.  1.    Given  log  2  =  .3010,  log  3  =  .4771 ;  find  log  .00432. 
We  have,  log  432  =  log  (24  x  33)  =  4  log  2  +  3  log  3  =  2.6353. 
Then  by  §  84,  the  mantissa  of  the  result  is  .6353. 
Whence  by  §  71,      log  .00432  =  7.6353  -  10. 

EXAMPLES. 
Given  log 2  =  .3010,  log 3  =  .4771,  log 7  =  .8451,  find: 

2.  log  3.6.  6.   log  .00343.  10.   log  .1944. 

3.  log  11.2.  7.   log  2880.  11.   log  202.5. 

4.  log  .84.  8.   log  .0392.  12.   log  1/6A. 

5.  log  .098.  9.   log  .000405.  13.   log  (14.7)1 

USE  OF  THE  TABLE  OF  LOGARITHMS   OF  NUMBERS. 

(For  directions  as  to  the  use  of  the  Table  of  Logarithms  of  Numbers, 
see  pages  iii  to  v  of  the  Introduction  to  the  Author's  Six  Place  Logarith- 
mic Tables.) 

EXAMPLES. 

86.  Find  the  logarithms  of  the  following  numbers : 

1.  .053.  5.  336.908.  9.  .001030746. 

2.  51.8.  6.  .000602851.  10.  .00000876092. 

3.  .2956.  7.  65000.63.  11.  730407.8. 

4.  1.027;4.  8.  9122.55.  12.  .0000436927. 

Find  the  numbers  corresponding  to  the  following  logarithms : 

13.  1.880814.  17.  8.044891-10.  21.   3.990191. 

14.  9.470410-10.         18.   2.270293.  22.   5.670180. 

15.  0.820204.  19.   7.350064-10.  23.   6.535003-10. 

16.  4.745126.  20.   5.000027-10.  24.   4.115658-10. 

APPLICATIONS. 

87.  The  approximate  value  of  ( an  arithmetical  quantity,  in  which  the 
operations  indicated  involve  only  multiplication,  division,  involution,  or 
evolution,  may  be  conveniently  found  by  logarithms. 

The  utility  of  the  process  consists  in  the  fact  that  addition  takes  the 
place  of  multiplication,  subtraction  of  division,  multiplication  of  invo- 
lution, and  division  of  evolution. 

Note.  In  computations  with  six-place  logarithms,  the  results  cannot  usually  be 
depended  upon  to  more  than  six  significant  figures. 


48  PLANE   TRIGONOMETRY. 

88.  1.   Find  the  value  of  .0631  x  7.208  x  .51272. 

By  §  75,  log  (.0631  x  7.208  x  .51272)  =  log  .0631  +  log  7.208  +  log  .51272. 

log  .0631=    8.800029-10 
log  7.208=   0.857815 
log  .51272=   9.709880-10 

Adding,    log  of  result  =  19.367724  -  20  =  9.367724  -  10.  (See  Note  1.) 
Number  corresponding  to  9.367724  —  10  =  .233197. 

Note  1.    If  the  sum  is  a  negative  logarithm,  it  should  be  written  in  such  a  form 
that  the  negative  portion  of  the  characteristic  may  be  —  10. 
Thus,  19.367724  -  20  is  written  in  the  form  9.367724  -  10. 

2.   Find  the  value  of  ||||. 

By  §  77'  log       =  log  336-852  ~  log  7980-04 


log  336.852  =  12.527439  -  10    (See  Note  2.) 
log  7980.04=   3.902005 

Subtracting,  log  of  result  =    8.625434  -  10 

Number  corresponding  =  .0422118. 

Note  2.  To  subtract  a  greater  logarithm  from  a  less,  or  to  subtract  a  negative 
logarithm  from  a  positive,  increase  the  characteristic  of  the  minuend  by  10,  writing 
—  10  after  the  mantissa  to  compensate. 

Thus,  to  subtract  3.902005  from  2.527439,  write  the  minuend  in  the  form 
12.527439  -  10  ;  subtracting  3.902005  from  this,  the  result  is  8.625434  -  10. 

3.   Find  the  value  of  (.0980937)5. 
By  §  79,  log  (.0980937)5  =  5  x  log  .0980937. 

log  .0980937  =  8.991641  -  10 


44.958205  -  50  =  4.958205  -  10.    (See  Note  1.) 
Number  corresponding  =  .0000090825. 

4.  Find  the  value  of  -v^.035063. 
By  §  80,  log  ^.035063  =  |  log  .035063. 

log  .035063  =    8.544849  -  10 

3)28.544849  -  30    (See  Note  3.) 
9.514950  -^10 

Number  corresponding  =  .327303. 


LOGARITHMS.  49 

Note  3.  To  divide  a  negative  logarithm,  write  it  in  such  a  form  that  the  negative 
portion  of  the  characteristic  may  be  exactly  divisible  by  the  divisor,  with  —  10  as  the 
quotient. 

Thus,  to  divide  8.544849  —  10  by  3,  we  write  the  logarithm  in  the  form 
28.644849  -  30  j  dividing  this  by  3,  the  quotient  is  9.514950  -  10. 

89.  Arithmetical  Complement. 

The  Arithmetical  Complement  of  the  logarithm  of  a  number,  or,  briefly, 
the  Cologarithm  of  the  number,  is  the  logarithm  of  the  reciprocal  of  that 
number. 

Thus.  colog  409  =  log  -i-  =  log  1  -  log  409. 

409 

log  1  =  10.  -  10  (Note  2,  §  88.) 

log  409  =   2.611723 

Then,  colog  409  =   7.388277  -  10. 

Again,  colog  .067  =  log  -i_  =  log  1  -  log  .067. 

.Uo7 

log  1  =  10.  -10 

log  .067  =    8.826075  -  10 

Then,  colog  .067=    1.173925. 

It  follows  from  the  above  that  the  cologarithm  of  a  number  may  be 
found  by  subtracting  its  logarithm  from  10  — 10. 

Note.  The  cologarithm  may  be  obtained  by  subtracting  the  last  significant 
figure  of  the  logarithm  from  10,  and  each  of  the  others  from  9,  —  10  being  written 
after  the  result  in  the  case  of  a  positive  logarithm. 

90.  Example.    Find  the  value  of       5L384 


8.709  x  .0946 

^^ 
=  log  51.384  +  log  -L-+  log 


8.709  .0946 

=  log  51.384  +  colog  8.709  -f  colog  .0946. 
log  51.384  =  1.710828 
colog  8.709  =  9.060032  -  10 
colog  .0946  =  1.024109 

1.794969  =  log  62.369. 

It  is  evident  from  the  above  example  that  the  logarithm  of  a  fraction 
whose  terms  are  composed  of  factors  may  be  found  by  the  following  rule : 

Add  together  the  logarithms  of  the  factors  of  the  numerator,  and  the 
cologarithms  of  the  factors  of  the  denominator. 


50 


PLANE  TRIGONOMETRY. 


Note.    The  value  of  the  above  fraction  may  be  found  without  using  cologarithms^ 
by  the  following  formula  : 

'  log51'384  -  log  (8'709  x  '0946) 

=  log  51  .384  -(log  8.709  +  log  .0946). 

The  advantage  in  the  use  of  cologarithms  is  that  the  written  work  of  computation 
is  exhibited  in  a  more  compact  form. 


EXAMPLES, 

Note.    A  negative  quantity  has  no  common  logarithm  (§  66,  Note). 

If  such  quantities  occur  in  computation,  they  should  be  treated  as  if  they  were 
positive,  and  the  sign  of  the  result  determined  irrespective  of  the  logarithmic  work. 

Thus,  in  Ex.  2,  §  91,  the  value  of  84.759  x  (-  2280.76)  is  obtained  by  finding  the 
value  of  84.759  x  2280.76,  and  putting  a  negative  sign  before  the  result.  See  also 
Ex.  29. 

91.  Find  by  logarithms  the  values  of  the  following  : 


1.  3.1425  x  603.93. 

2.  84.759  x  (-2280.76). 


5 

' 


9 


4867.2 


1.05478 


765.16  '   34.9564 

3.89612  x  .6946 


3.  (-4.39182)  x  (-  .0703968). 

4.  .936537  x  .00117854. 

-  2.7085  g     -  .000680239 

'  .0868097*  .00512643 

.-          (-  .870284)  x  3.73 

'  (-  .06585)  x  (-42.317)' 

12  .082136  x  (-  73.39) 
.838  x  2808.72 


23.    -\/100. 


25.    ^.0725628. 


4694.9  x  .00454 
1Q      715  x  (-  .024158) 
'    (-  .5157)  x  1420.63* 

13.  (7.7954)4. 

14.  (.83287)7. 

15.  (-25.1437)3. 

16.  (.01)*. 

17.  (-964.38)*. 

28.   Find  the  value  of 
By  §  90, 

log?^?  =  Iog2  +  log 
3* 

log  2=    .301030 

log  5  =    .698970  ;  divide  by  3  =    .232990 

colog  3  =  9.522879  -  10  ;  multiply  by  f  =  9.602399  -  10 


18.  (.0951293)1 

19.  (.000105936)1 

20.  V§. 

21.  -\/2. 

22.  -ty~^6. 


26.  -X/.002613874. 

27.  ^-.000951735. 


3* 


+  colog3*  =  Iog2  +  ±  Iog5  +  |  colog3. 


.136419  =  log  1.36905, 


LOGARITHMS. 


51 


29.    Find  the  value 

'Hi = * log  £iii = *  **  -032956  - log  7-96183>- 

log  .032956  =  8.517934  -  10 
log  7.96183  =  0.901013 

3)27.616921  -  30 

9.205640  -  10  =  log  .160561. 
Eesult,  -.160561. 
Find  the  values  of  the  following : 

4400  \$  40. 


6.1  x  . 


1.3073      ) 

75.431T" 
31.4  x  .4146* 

A/.000965782 


V.00497836 

44      -  (.256929)*. 

(-.834574)* 
(.5732)* 


V5106.526  x  .000031093. 

47.  (837.48  x  .00943246)'. 

48.  (4.867184)^  x  (.175437)^ 
V3.9285  x 


8693.84  xV.033074 

(-  .00019162)^  x  V6818 
-  .2755653 


vC374xV.007835912 

EXPONENTIAL  EQUATIONS. 

92.   An  Exponential  Equation  is  an  equation  of  the  form  a*  =  b. 

To  solve  an  equation  of  this  form,  take  the  logarithms  of  both  mem* 


bers. 


1.   Given  31*  =  23 ;  find  the  value  of  x. 
Taking  the  logarithms  of  both  members, 

log  (31*)  =  log  23. 

Whence  by  §  79,  x  log  31  =  log  23. 

=  log  23  _ 
log  31 = 


Then, 


1.361728 
1.491362 


=  .9130  +. 


52  PLANE   TRIGONOMETRY. 

2.   Given  .2*  =  3  ;  find  the  value  of  x. 
Taking  the  logarithms  of  both  members, 

a;  log  .2  =  log  3. 


Whence,  *  =         >=       -477121       =    -477121    =  -.6826 
log  .2     9.301030  -  10      -  .698970 


EXAMPLES. 
Solve  the  following  equations  : 

3.  332.9*  =  5.178.  5.   .0158*  =  .0082958.  7.  a*  = 

4.  .4162*  =  6.724.  6.   5.3364"  =  .744.  8.  m2a*  =  n4. 

9.   62s-3=.  0277778.  10.   .7"*+*  =  .16807. 

93.  1.   Find  the  logarithm  of  .3  to  the  base  7. 


EXAMPLES. 

Find  the  values  of  the  following  : 

2.  loga  13.  4.   log.746.2.  6.  Iog9>1.362. 

3.  Iog5  .9.  5.  log.48.087.  7.   log®  4.3. 

Examples  like  the  above  may  be  solved  by  inspection  if  the  numbei 
can  be  expressed  as  an  exact  power  of  the  base. 

8.  Find  the  logarithm  of  128  to  the  base  16. 
Let  logw  128  ==  x;  then  by  §  66,  16X  =  128. 
That  is,  (24)*  =  27,  or  24<°  =  27. 

*r 

Whence  by  inspection,  4  x  =  7  ;  and  x  =  Iog16  128  =  -• 

4 

9.  Find  the  logarithm  of  81  to  the  base  3. 

10.  Find  the  logarithm  of  32  to  the  base  8. 

11.  Find  the  logarithm  of  ^  to  the  base  27. 

12.  Find  the  logarithm  of  ^  to  the  base  -3^. 


LOGARITHMS.  53 

EXAMPLES  IN  THE  USE  OF  TRIGONOMETRIC  TABLES. 

(For  directions,  see  pages  v  to  xi  of  the  Introduction  to  the  Author's 
Six  Place  Logarithmic  Tables.)  ¥ 

94.  Table  of  Logarithmic  Sines,  Cosines,  etc. 

Find  the  values  of  the  following : 

1.  log  sin  12°  48' 52".        4.   log  cot  53°  42 '9".  7.   log  cot  26°  30 '14". 

2.  Iogtan67°13'27".        5.   log  cos  79°  54 '35".  8.   log  sec  45°  26 '38". 

3.  Iogcos31°5'43".          6.   log  tan 8°  17 '21".  9.   log  esc  84° 9 '56". 

Find  the  angles  corresponding  in  the  following : 

10.  log  sin  =  9.934232  -  10.  14.   log  tan  =  9.184367  -  10. 

11.  log  cos  =  9.923569  -  10.  15.   log  cot  =  9.404692  -  10. 

12.  log  tan  =0.806571.  16.  log  sec  =  0.188783. 

13.  log  cot  =0.282956.  17.   log  esc  =  0.400314. 

95.  Table  of  Natural  Sines,  Cosines,  etc. 

Find  the  values  of  the  following : 

1.  sin43°17'35".  3.   cos  86° 21 '46".  5.  sin67d9'54". 

2.  cot75°50'19".  4.  tan34°48'23".  6.  cos29°35'8". 

Find  the  angles  corresponding  in  the  following : 
7.  tan  =  1.2622.      8.  cos  =  .96376.     9.  sin  =  .91527.      10.  cot  =  1.7927 

96.  Auxiliary  Table  for  Small  Angles. 

Find  the  values  of  the  following : 
1.  log  sin  1°  14' 53".  2.   log  tan  3°  42 '8".  3.  log  cot  2°  26 '35". 

Find  the  angles  corresponding  in  the  following : 
4.  log  sin  =  8.233459 -10.  5.   log  tan  =  7.859872  - 10. 

6.   log  cot  =  1.546267. 


54  PLANE  TRIGONOMETRY. 


VI.    SOLUTION  OP  RIGHT  TRIANGLES. 

97.  The  elements  of  a  triangle  are  its  three  sides  and  its  three  angles. 

We  know  by  Geometry  that  a  triangle  is,  in  general,  completely  deter- 
mined when  three  of  its  elements  are  known,  provided  one  of  them  is  a 
side. 

The  solution  of  a  triangle  is  the  process  of  computing  the  unknown 
from  the  given  elements. 

98.  To  solve  a  right  triangle,  two  elements  must  be  given  in  addition 
to  the  right  angle,  one  of  which  must  be  a  side. 

The  various  cases  which  can  occur  may  all  be  solved  by  aid  of  the 
following  formulae : 


.     A      a  A      b  ,       A      a 

sin  A  =  —  cos  ^1  =  —  tan  A.  =  •=•  • 

ceo 


sin  5  =  —  cos  5  =  — 

o  c  -a 

99.   CASE  I.    When  the  given  elements  are  a  side  and  an  angle. 

The  proper  formula  for  computing  either  of  the  remaining  sides  may 
be  found  by  the  following  rule : 

Take  that  function  of  the  angle  which  involves  the  given  side  and  the 
required  side. 

1.    Given  c  =  68,  B  =  21°  42' 39";  find  a  and  6. 
In  this  case  the  formulas  to  be  used  are 

a  b 

cos  B  =  -,  and  sin  B  =  — 
c  c 

Whence,  a  —  c  cos  B,  and  b  =  c  sin  B.  (A) 

Solution  by  Natural  Functions. 
a  =  68  x  cos 21° 42' 39"  =  68  x  .92906  =  63.176. 
6  =  68  x  sin  21°  42' 39"  =  68  x  .36993  =  25.155. 


SOLUTION  OF  RIGHT  TRIANGLES.  55 

Solution  by  Logarithms. 
Taking  the  logarithms  of  both  members,  in  formulae  (A), 

log  a  =  log  c  +  log  cos  B,  and  log  b  =  log  c  +  log  sin  B. 

log  c  =  1.832509  log  c  =  1.832509 

log  cos  B  =  9.968045  -  10  log  sin  B  =  9.568111  -  10 

log  a  =  1.800554  log  b  =  1.400620 

a  =  63.1762.  6  =  25.1547. 

2.   Given  a  =  .235867,  A  =  67°9'23";  find  b  and  c. 

In  this  case,  tan  A  =  ^,  and  sin  A  =  -• 

b  c 

Whence,  b  =  —^-7,  and  c  =  -r-^-r 

tan  A  sin  A 

By  logarithms,  log  b  —  log  a  —  log  tan  A,  and  log  c  =  log  a  —  log  sin  A. 

log  a  =  9.372667  -  10  log  a  =  9.372667  - 10 

log  tan  A  =  0.375452  log  sin  A  =  9.964527  -  10 

log  b  =  8.997215  -  10  log  c  =  9.408140  -  10 

b  =  .0993607.  c  =  .255941. 

100.  CASE  II.    When  both  the  given  elements  are  sides. 

First  calculate  one  of  the  angles  by  aid  of  either  formula  involving 
the  given  elements,  and  then  compute  the  remaining  side  by  the  rule  of 
Case  I. 

Example.     Given  b  -  .15124,  c  =  .30807 ;  find  A  and  a. 

5"f  ormula  cos  A  =  -,  and  then  find  a  by  the 

formula  sin  A  =  — .  or  a  =  c  sin  A. 
c 

By  logarithms,  log  cos  A  =  log  6  —  log  c,  and  log  a  —  log  c  +  log  sin  A. 

log  6  =  9.179667  - 10  log  c  =  9.488650  - 10 

log  c  =  9.488650  -  10  log  sin  A  =  9.940118  -  10 

log  cos  A  =  9.691017  -  10  log  a  =  9.428768  -  10 

A  =  60°  35'  54.4".  a  =  .268391. 

101.  In  the  Trigonometric  solution  of  any  example  under  Case  II., 
it  is  r»  -eessary  to  first  find  one  of  the  angles,  and  the  remaining  side  may 
then   )e  calculated. 

I  possible,  however,  to  compute  the  third  side  directly,  without  f  ;  ' 
finding  the  angle,  by  Geometry. 


56  PLANE  TRIGONOMETRY. 

Thus,  in  the  example  of  §  100,  we  have,  by  Geometry,  a2  +  62  =  c2. 


Whence,  a  =  Vc2-^2  =  V(c  +  6)(c-6). 

By  logarithms,  log  a  =  -J-  [log  (c  -f-  6)  +  log  (c  —  6)]. 

c  +  b  =  .45931  ;  log  =  9.662106  -  10 
c  -  b  =  .15683  ;  log  =  9.195429  -  10 

2)18.857535  -  20 
log  a  =  9.428768  -10 
a  =  .268391,  as  before. 

If  the  given  sides  are  a  and  6,  the  expression  for  c  is  Va2  4-  b2,  which 
is  not  adapted  to  logarithmic  computation. 

In  such  a  case,  it  is  usually  shorter  to  proceed  as  in  §  100. 

EXAMPLES. 

Note.  In  those  examples  of  the  following  set  in  which  the  given  sides  are  num- 
bers of  not  more  than  three  significant  figures,  and  the  operations  indicated  involve 
only  multiplication,  it  is  usually  shorter  to  employ  Natural  Functions. 

In  such  a  case,  the  results  cannot  be  depended  upon  to  more  than  five  significant 
figures  ;  while  in  the  solutions  by  logarithms,  they  can  be  depended  upon  to  six  signifi- 
cant figures. 

102.   Solve  the  following  right  triangles  : 

1.  Given  A  =  16°,   c  =  7.  9.  Given  ^  =  9°,         6  =  937. 

2.  Given  B  =  67°,  a  =  5.  10.  Given  a  =  3.414,  6  =  2.875.      V 

3.  Given  B  =  50°,  6  =  20.  11.  Given  ^1  =  84°  16',  a  =  .0033503. 

4.  Given  a  =  .35,    c=.62.  12.  Given  A  =  46°  23',  c  =  5278.6. 

5.  Given  a  =  273,  6  =  418.  13.  Given  a  =  529.3,   c  =  902.7. 

6.  Given  ^  =  38°,  a  =  8.09.  14.  Given  B  =  23°  9  ',  6  =  75.48. 

7.  Given  B  =  75°,   c  =  .014.  15.  Given  A  =  72°  52',  6  =  6306. 

8.  Given   6  =  58.6,  c  =  76.3.  16.  Given  B  =  18°  38',  c  =  2.5432. 

17.  Given   a  =.0001689,      6  =  .0004761. 

18.  Given  ^1  =  31°  45',        a  =  48.0408. 

19.  Given  6  =  617.57,          c  =  729.59. 

20.  .  Given  B  =  82°  6'  18",    a  =  89.32. 

21.  Given  A  =  55°  43'  29",  c  =  41518. 

22.  Given  B  =  31°  47'  7",    a  =  7.23246. 

23.  Given  a  =  99.464,         c  =  156.819. 


SOLUTION  OF  RIGHT  TRIANGLES.  57 

24.  Given  A  =  43°  21'  36",  b  =  .00261751. 

25.  Given  B  =  79°  14'  31",  b  =  84218.5. 

26.  Given  B  =  67°  39'  53",  c  =  9537514. 

27.  Given  6  =  5789.72,        c  =  24916.45. 

28.  Given  A  =  26°  12'  24",  c  =  469422.7. 

29.  Given  B  =  14°  55'  42",  b  =  .1353371. 

30.  Given  a  =  672.3853,      6  =  384.5038. 

Solve  the  following  isosceles  triangles,  in  which  A  and  B  are  the  equal 
angles,  and  a,  6,  and  c  the  sides  opposite  the  angles  A,  B,  and  (7,  respec- 
tively : 

31.  Given  .4  =  68°  57',         6  =  350.94. 

32.  Given  B  =  27°  8',  c  =  3.0892. 

33.  Given  C  =  84°  47',         6  =  91032.7. 

34.  Given  a  =  79.2434,        c=  106.6362. 

35.  Given  A  =  35°  19'  47",  c  =  .56235. 

36.  Given  0=  151°  28' 52",  c  =  9547.12. 

37.  A  regular  pentagon  is  inscribed  in  a  circle  whose  diameter  is  35. 
Find  the  length  of  its  side. 

38.  At  a  distance  of  105  ft.  from  the  base  of  a  tower,  the  angle  of 
elevation  of  its  top  is  observed  to  be  38°  25'.     Find  its  height. 

39.  What  is  the  angle  of  elevation  of  the  sun  when  a  tower  whose 
height  is  103.74  ft.  casts  a  shadow  167.38  ft.  in  length  ? 

40.  If  the  diameter  of  a  circle  is  32689,  find  the  angle  at  the  centre 
subtended  by  an  arc  whose  chord  is  10273. 

41.  If  the  diameter  of  the  earth  is  7912  miles,  what  is  the  distance  of 
the  remotest  point  of  the  surface  visible  from  the  summit  of  a  mountain 
1  \  miles  in  height  ? 

42.  Find  the  length  of  the  diagonal  of  a  regular  pentagon  whose  side 
is  6.3257. 

43.  Find  the  angle  of  elevation  of  a  mountain-slope  which  rises  238  ft. 
in  a  horizontal  distance  of  one-eighth  of  a  mile. 

44.  From  the  top  of  a  lighthouse,  146  ft.  above  the  sea,  the  angle  of 
depression  of  a  buoy  is  observed  to  be  21°  46'.     Find  the  horizontal  dis- 
tance of  the  buoy. 


•• 


58  PLANE   TRIGONOMETRY. 

45.  If  a  pole  casts  a  shadow  which  is  two-thirds  its  own  length,  what 
is  the  angle  of  elevation  of  the  sun  ? 

46.  A  vessel  is  sailing  due  east  at  the  rate  of  7.8  miles  an  hour.     A 
headland  is  observed  to  bear  due  north  at  10.37  A.M.,  and  33°  west  of 
north  at  12.43  P.M.     Find  the  distance  of  the  headland  from  each  point  of 
observation. 

47.  If  a  chord  whose  length  is  41.368  subtends  an  arc  of  145°  37',  what 
is  the  radius  of  the  circle  ? 

48.  The  length  of  the  side  of  a  regular  octagon  is  12.     Find  the  radii 
of  the  inscribed  and  circumscribed  circles. 

49.  How  far  from  the  foot  of  a  flagpole  110  ft.  high  must  an  observer 
stand,  so  that  the  angle  of  elevation  of  the  top  of  the  pole  may  be  12°  ? 

50.  If  the  diagonal  of  a  regular  pentagon  is  32.835,  what  is  the  radius 
of  the  circumscribed  circle  ? 

51.  From  the  top  of  a  tower,  the  angle  of  depression  of  the  extremity 
of  a  horizontal  base  line,  1250  ft.  in  length  measured  from  the  foot  of  the 
tower,  is  observed  to  be  18°  36' 29".     Find  the  height  of  the  tower. 

52.  If  the  radius  of  a  circle  is  723.294,  what  is  the  length  of  a  chord 
which  subtends  an  arc  of  35°  13 '  ? 

53.  A  regular  hexagon  is  circumscribed  about  a  circle  whose  diameter 
is  18.     Find  the  length  of  its  side. 

54.  From  the  top  of  a  lighthouse  200  ft.  above  the  sea,  the  angles  of 
depression  of  two  boats  in  line  with  the  lighthouse  are  observed  to  be  14° 
and  32°,  respectively.     Find  the  distance  between  the  boats. 

55.  A  vessel  is  sailing  due  east  at  a  uniform  rate  of  speed.     At  7  A.M., 
a  lighthouse  is  observed  bearing  due  north,  10.326  miles  distant ;  and  at 
7.30  A.M.  it  bears  18°  13'  west  of  north.     Find  the  rate  of  sailing  of  the 
vessel,  and  the  bearing  of  the  lighthouse  at  10  A.M. 

103.  Care  must  be  taken  to  use  the  Auxiliary  Table  for  Small  Angles 
in  finding  the  logarithmic  functions  of  angles  between  0°  and  5°,  or  between 
85°  and  90°,  or  the  angles  corresponding  in  the  same  cases. 

This  provides  for  every  case  which  can  arise  in  solving  right  triangles, 
except  in  looking  out  the  angle  corresponding  to  a  logarithmic  sine  when 
between  85°  and  90°,  or  a  logarithmic  cosine  when  between  0°  and  5°. 

We  will  now  derive  a  formula  for  right  triangles  by  aid  of  which, 
when  b  and  c  are  given,  the  angle  A  may  be  determined  with  accuracy 
if  it  is  between  85°  and  90°. 


J 


SOLUTION  OF  RIGHT  TRIANGLES.  59 

By  §98,  cosA  =  -. 

Then  by  (31),  2  sin2  \A  =  1  -  cosA=  1  -  -  =  ^-^ 

c        c 

i r 

Therefore,  sin  ^  A  = 


In  like  manner,  sin -J- B  =  A/C      a« 

These  formulae  involve  the  half-angles;  hence,  if  the  angle  itself  is 
between  85°  and  90°,  its  half  is  between  42°  30'  and  45°,  and  the  cor- 
rection in  seconds  may  in  that  case  be  found  from  the  table  with  sufficient 
.precision. 

An  angle  between  0°  and  5°  may  always  be  avoided  in  solving  a  right 
triangle  by  working  with  the  other  acute  angle. 

104.  1.  Given  6  =  1.08249,  c  =  1.08261 ;  find  the  angles. 
Here  A  is  near  to  0°,  and  B  is  near  to  90°,  as  may  be  determined  by 
inspection. 

We  then  proceed  to  find  B  by  the  formula  of  §  103. 

For  this  purpose,  we  must  first  find  a,  which  may  be  done  as  in  §  101 . 

c  +  b  =  2.1651 ;  log  =  0.335478 

c  -  b  =  .00012 ;  log  =  6.079181  -  10 

2)16.414659  -~20 
log  a  =  8.207330 -10 

Whence  a  =.0161187. 

Now  to  find  B,  we  use  the  formula  sin  \B  =\—e  —• 

*     Zi  C 

By  logarithms,  log  sin \B  =  | [log (c  -  a) -  log 2 c] . 

c  -  a  =  1.0664913 ;  log  =  0.027957 
2  c  =  2.16522 ;      log  =  0.335502 

2)19.692455  -"20 
log  sin  i  B  =  9.846228  -  10 

Whence,  ££  =  44°  34'  24.7". 

Then,     B  =  89°  8'  49.4",  and  A  =  90°  -  B  =  0°  51'  10.6". 

If  b  is  small  compared  with  c,  then  A  is  near  to  90°,  and  should  be 
calculated  directly  by  aid  of  the  formula  of  §  103. 


60  PLANE   TRIGONOMETRY. 

EXAMPLES. 
In  each  of  the  following  right  triangles  find  the  angles : 

2.  Given  a  =  .0128,       c  =  152.337. 

3.  Given  6  =  5.81006,  c  =  5.81039. 

4.  Given  c  =  11527.2,  6  =  1.32. 

5.  Given  a  =  .77,  c  =  98276.4. 

6.  Given  a  =  42.0098,  c  =  42.0103. 

FORMULAE  FOR  THE  AREA  OF  A  RIGHT  TRIANGLE. 
105.  CASE  I.     Given  the  hypotenuse  and  an  acute  angle. 

B 


b 
Denoting  the  area  by  K,  we  have  by  Geometry, 


But  by  §  4,  a  =  c  sin  A,  and  b  =  c  cos  A. 

Whence,  •       2  K=  c2  sinJ.  cos  A  =  |  c2  sin  2  -4/by.  (25). 
Then,  4  K=  c2  sin  2  A.  (38) 

In  like  manner,  4  K  =  c2  sin  2  B.  (39) 

CASE  II.     Given  an  angle  and  its  opposite  side. 

By  §4,  5  =  acotA 

Whence,  2  K  =  a  x  a  cot  A  =  a2  cot  A.  (40) 

In  like  manner,  2K=l2 cot #.  (41) 

CASE  III.     Given  an  angle  and  its  adjacent  side. 

By  §  4,  6  =  atan.B. 

Whence,  2  K  =  a  x  a  tan  £  =  a2  tan  5.  (42) 

In  like  manner,  2  .ST  =  62  tan  A  (43) 


SOLUTION  OF  RIGHT  TRIANGLES.  61 

CASE  IV.     Given  the  hypotenuse  and  another  side. 
By  Geometry,  b2  =  c2  —  a2. 


Whence,        2  K=  ab  =  a  Vc2  —  a2  =  a  V(c  +  a)(c  —  a).  (44) 


In  like  manner,  2  K=  b  V(c  +  6)  (c  —  6).  (45) 

CASE  V.     (riven  Zfte  taco  sides  a&ow£  £fte  right  angle. 

In  this  case,  2K=ab.  (46) 

EXAMPLES. 

106.  1.   Given  c  =  10.3572,  5  =  74°  57'  14";  find  the  area. 
By  (39),  4  K  =  c2  sin  2  jB. 

Whence,  log  (4  JT)  =  2  log  c  +  log  sin  2  5. 

log  c  =  1.015242 ;  multiply  by  2  =  2.030484 
2  B  =  149°  54'  28  " ;          log  sin  =  9.700178  - 10 

log  (4  K)  =  1.730662 

4^=53.7851 
Dividing  by  4,  K=  13.4463. 

Note.    To  find  log  sin  149°  54'  28' ',  take  either  log  cos  59°  54'  28",  or  log  sin  30C 
5'  32".     (See  Introduction  to  Tables,  page  viii.) 

Find  the  areas  of  the  following  right  angles : 

2.  Given  A  =  19°  36',  a  =  2.2178.       4.   Given  a  =  149.417,  6  =  76.292. 

3.  Given  B  =  24° 7 '48",  a  =  .8213.     5.   Given  b  =  .305694,  c=. 660156. 

6.  Given  .4  =  30°  56'  19",  c  =  192.035. 

7.  Given  .4=78°  42'  53",  6  =  .0520281. 

8.  Given  a  =  .932368,  c  =  4.786723. 

9.  Given  B  =  72°  18'  27",  c  =  27.28338. 
10.  Given  B  =  49°  25'  34",  &  =  . 3375494. 


PLANE  TRIGONOMETRY. 


VII.    GENERAL  PROPERTIES  OP  TRIANGLES. 

107.   In  any  triangle,  the  sides  are  proportional  to  the  sines  of  their 
opposite  angles. 

I.   To  prove  a  :  b  =  sin  A :  sin  B.  (47) 

C 


c        JD 
FIG.  1.       ' 


D 


There  will  be  two  cases,  according  as  the  angles  A  and  B  are  both 
acute  (Fig.  1),  or  one  of  them  obtuse  (Fig.  2). 
In  each  case,  draw  CD  perpendicular  to  AB. 
Then  in  each  figure,  CD  =  b  sin  A  (§  4). 

Also  in  Fig.  1,  CD  =  a  sin  B. 

And  in  Fig.  2,  CD  =  a  sin  CBD 

=  a  sin  (180°  -  JB)  =  a  sin  B  (§  33). 
b  sin  A  =  a  sin  B. 


Then  in  either  case, 

Whence  by  the  theory  of  proportion, 

a  :  b  =  sin  A  :  sin  B. 

In  like  manner,   '  b  :  c  =  sin  B :  sin  (7, 

and  c :  a  =  sin  C :  sin  A. 


(48) 
(49) 


108.   Jw  any  triangle,  the  sum  of  any  two  sides  is  to  their  difference  as 
the  tangent  of  half  the  sum  of  the  opposite  angles  is  to  the  tangent  of  half  their 
Terence. 
By  (47),  a  :  b  =  sin  A  :  sin  B. 

Whence  by  composition  and  division, 

a  +  b  :  a  —  b  =  sin  A  +  sin  B :  sin  A  —  sin  B. 


Or, 
But, 


a  +  b  __  sin  J.  +  sin  B 
a  —  b      sin  ^4  —  sin  B 

5 

sin  .4  +  sin  B_  tan-fr  (J.  +  ff)   hv 
~  '    * 


sin  ^  -  sin  jB  ~  tan 


GENERAL  PROPERTIES  OF  TRIANGLES.  63 


-        . 

Whence.  !IT  =  ^    f-hr  ~w:'  (50) 

a  —  b     tan  ^  (  A  —  B) 


109.  Jw  any  triangle,  the  square  of  any  side  is  equal  to  the  sum  of  the 
squares  of  the  other  two  sides,  minus  twice  their  product  into  the  cosine  of 
their  included  angle. 

I.   To  prove  a2  =  b2  -f  c2  —  2  be  cos  A.  (53) 

CASE  I.     When  the  included  angle  A  is  acute.     (Figures  of  §  107.) 

There  will  be  two  cases,  according  as  the  angle  B  is  acute  (Fig.  1),  or 
obtuse  (Fig.  2). 

Then  in  Fig.  1,  BD  =  c  -  AD,  and  in  Fig.  2,  BD  =  AD  -  c. 
Squaring,  we  have  in  either  case, 

Bff  =  AD2  +  c2  -  2  c  x  AD. 
Adding  CD2  to  both  members, 

BD2  +  CZ?  =  AD2  +CD2  +  c2-2cx  AD. 
But  BD*+CI?  =  a2,  and  ZD2 

Also,  by  §  4,  AD  =  6  cos  A. 

Whence,  a2  =  62  -f  c2  —  2  be  cos  -4. 

CASE  II.    When  the  included  angle  A  is  obtuse. 


B       CAD 

Draw  CD  perpendicular  to  AB. 

We  have  BD  =  AD  +  c. 

Squaring,  and  adding  CD2  to  both  members, 


64  PLANE  TRIGONOMETRY. 

But  B&  +  CI?=a2,  and  ZZ?  +  C^2  =  b2. 

And  by  §  4,     AD  =  b  cos  CAD  =  b  cos  (180°  -  ^)  =  -  b  cos  J.  (§  33). 
Whence,  a2  =  b2  +  c2  -  2  6c  cos  A 

In  like  manner,  b2  =  c2  +  a2  —  2  ca  cos  5,  (54) 

and  c2  =  a2  +  b2  —  2  ab  cos  (7.  (55) 

110.     To  express  the  cosines  of  the  angles  of  a  triangle  in  terms  of  the 
sides  of  the  triangle. 

By  (53),  a2  =  b2  +  c2  -  2  be  cos  A. 

Transposing,  2  be  cos  A '  —  b2  +  c2  —  a2. 


Whence,  cos  J.  = "  "V         •  (56) 


In  like  manner,  cos  B  =  -~^ —>  (57) 

and  COsO  =  ^^.  (58) 

111.    To  express  the  sines,  cosines,  and  tangents  of  the  half-angles  of  a 
triangle  in  terms  of  the  sides  of  the  triangle. 

By  (56),  1 


2  be  2  be 


Whence,  by  (31),    2  sin2  £  ^  =  ^    2  6<T 


Or,  sin2  £.4  = 


46c 
Denoting  the  sum  of  the  sides,  a  +  b  4-  c,  by  2  s,  we  have 


Whence,  si: 


Or,  sin^=^         £   •      '• 


GENERAL  PROPERTIES  OF  TRIANGLES. 


65 


In  like  manner,  sin  1 B  = 


and 


ao 


Again,by(56), 


+  cos  A  = 


Whence,  by  (32),    2  cos2  J  J.  =         <?- 

^  oc 


Or, 
But, 


cos2 1  ^  = 


and 


6  -j-  c  -  a  =  (6  +  c  +  a)  —  2  a  =  2  (s  —  a). 


Whence, 

Or, 

In  like  manner, 


and 


Dividing  (59)  by  (62),  we  have,  by  (4), 


tan 


In  like  manner, 


tan 


**°V(-K?Vj 


and 


s(s  —  c) 


(60) 


(61) 


(62) 
(63) 
(64) 

(65) 
(66) 
(67) 


Note.    Since  each  angle  of  a  triangle  is  less  than  180°,  its  half  is  less  than  90°  ; 
hence,  the  positive  sign  must  be  taken  before  the  radical  in  each  formula  of  §  111. 

FORMULA  FOR  THE  AREA  OF  AN  OBLIQUE  TRIANGLE. 

112.   CASE  I.     Given  two  sides  and  their  included  angle. 
I.  When  the  given  parts  are  6,  c,  and  A. 


66 


PLANE  TRIGONOMETRY. 
C 


D 


FIG.  1. 


There  will  be  two  cases,  according  as  A  is  acute  (Fig.  1),  or  obtuse 
(Fig.  2). 

In  each  case,  draw  CD  perpendicular  to  AB. 

Then  denoting  the  area  by  K,  we  have  by  Geometry, 

2  K=  c  x  CD. 

.But  in  Fig.  1,  CD  =  b  sin  A  (§  4). 

And  in  Fig.  2,  CD  =  b  sin  CAD 

=  b  sin  (180°  -  A)  =  b  sin  A  (§  33). 

Then  in  either  case,       2  K=  be  sin  A  (68) 

In  like  manner,  2  K=ca  sin  B,  (69) 

and  2K=ab  sin  O.  (70) 

CASE  II.     Given  one  side  and  all  the  angles. 
I.   When  the  given  parts  are  a,  ^4,  .B,  and  C. 
By  (70),  2  K=  ab  sin  O. 

But  by  (47), 


6  _  sin  B         ,  _  a  sin  B 
a     sin  ^4.'  sin  A 


Whence, 

In  like  manner, 


=  a  x 


X  sin  C  =  — 


sin  B  sin 


sin  A  sin  A 

W  sin  C  sin  .4 


rr 
K  — 


and 


sin  B 

c2  sin  J.  sin  B 
sin  (7 


CASE  III.     Given  the  three  sides. 

By  (68),     2  K=  be  sin  J.  =  2  6c  sin  £  A  cos  ^  -4  by  (25). 

Dividing  by  2,  and  substituting  the  values  of  sin-i-J.  and 
from  (59)  and  (62),  we  have, 


(71) 
(72) 
(73) 


7T= 


=  V«(.-a)(a-6X.-c).       (74) 


SOLUTION  OF  OBLIQUE  TRIANGLES.  67 


VIII.    SOLUTION  OF  OBLIQUE  TRIANGLES. 

113.  In  the  solution  of  oblique  triangles,  we  may  distinguish  four 
eases*. 

114.  CASE  I.     Given  a  side  and  any  two  angles. 

The  third  angle  may  be  found  by  Geometry,  and  then  by  aid  of  §  107 
the  remaining  sides  may  be  calculated. 

The  triangle  is  always  possible  for  any  values  of  the  given  elements, 
provided  the  sum  of  the  given  angles  is  less  than  180°. 

1.   Given  b  =  20.24,  A  =  103°  36',  B  =  19°  21'  ;  find  <7,  a,  and  c. 
We  have,  C  =  180°  -(A  +  B)=  180°  -  122°  57'  =  57°  3'. 


-Rv/47N  _  ,   c  _  sin  C 

By(47)>  ~a    l      ~ 


Then,  a  —  b  sin  A  esc  B,  and  c  =  b  sin  C  esc  B. 

Whence,  log  a  =  log  b  +  log  sin  A  +  log  esc  B, 

and  log  c  =  log  b  +  log  sin  C  +  log  esc  B. 

log  b  =  1.306211  log  b  =  1.306211 

log  sin  A  =  9.987649  -  10  log  sin  C  =  9.923837  -  10 

log  esc  B  =  0.479729  log  esc  B  =  0.479729 

log  a  =  1.773589  log  c  =  1.709777 

a  =  59.3730.  c=  51.2598. 

Note.  To  find  log  sin  103°  36',  take  either  log  cos  13°  36',  or  log  sin  76°  24'.  To 
find  the  log  cosecant  of  an  angle,  subtract  the  log  sine  from  10  —  10.  (See  Introduc- 
tion to  Tables,  page  viii.) 

EXAMPLES, 
Solve  the  following  triangles  : 

2.  Given  a  =  180,  ^1  =  38°,  B  =  75°  43'. 

3.  Given  b  =  .82,     B  =  51°  42'  37",  C  =  109°  17'  23". 

4.  Given  c  =  24.637,   .4  =  83°  39',   *jB  =  38°56'. 

5.  Given  6  =  .06708,      ^  =  26°  10'  45",  0=44°  35'  12". 

6.  Given  a  =  5.0454,      B  =98°  8'  26",     O  =  21°  51'  34". 

7.  Given  c  =  4592.36,    J.=  74°27',          (7=61°. 

8.  Given  c  =  .93109,      A  =  15°  34'  9",     C  =  123°  29'  46". 

9.  Given  b  =  3.67683,    A  =  67°  21'  54",  B  =  57°  48'  8". 
10.   Given  a  =  71396.72,  .B  =  42°  55'  13",   C=16°4'57". 


68  PLANE   TRIGONOMETRY. 

115.   CASE  II.     Given  two  sides  and  their  included  angle. 

Since  one  angle  is  known,  the  sum  of  the  other  two  angles  may  be 
found  ;  and  then  their  difference  may  be  calculated  by  aid  of  §  108. 

Knowing  the  sum  and  difference  of  the  angles,  the  angles  themselves 
may  be  found;  and  then  the  remaining  side  may  be  computed  as  in 
Case  I. 

The  triangle  is  possible  for  any  values  of  the  data. 

1.   Given  a  =  82,  c  =  167,  B  =  98°  14';  find  A,  C,  and  6. 
By  Geometry,  C  +  A  =  180°  -  B  =  81°  46'. 


Or, 

C  ~Y  ^ 

Then,  log  tan  £  (C  -  A)  =  log  (c  -  a)  +  colog  (c  +  a)  +  log  tan  £  (C  +  A). 

c  -  a  =  85  log  =  1.929419 

c  +  a  =  249  colog  =  7.603801  -  10 

A)=  40°  53'  log  tan  =  9.937377  -  10 


log  tan  ±(C-A)  ='9.470597  -  10 
J((7-  4)=  16°  27'  49.8". 

Therefore,  C  =  %(C  +  A)+%(C  -  A)=  57°  20'  49.8", 

and  A  =  £  (C  +  ^)  -  £  (0  -  ^)  =  24°  25'  10.2". 

To  find  the  remaining  side,  we  have  by  (47), 

sin 
sin  .4 
Whence,  log  b  =  log  a  +  log  sin  B  +  log  esc  A. 

log  a  =  1.913814 
log  sin  £  =  9.995501  -10 
log  esc  ^  =  0.383615 

log  6  =  2.292930 
b  =  196.305. 

EXAMPLES, 
Solve  the  following  triangles  : 

2.  Given  a  =  67,  c  =  33,  5  =  36°. 

3.  Given  a  =  886,  6  =  747,  C  =  71°54'. 

4.  Given  6  =  4.102,        c  =  4.549,        .4=62°  9'  38". 


SOLUTION  OF  OBLIQUE   TRIANGLES.  69 

5.  Given  a  =  .5953,        b  =  .9639,        0  =  134°. 

6.  Given  b  =  1292.1,      c  =  286.3,        ^.  =  27°  13'. 

7.  Given  a  =  7.48,          c  =  12.409,      5  =  83°  26' 52". 

8.  £iven  a  =  93.273,      6  =  81.612,       (7  =  58°. 

9.  Given  b  =  .0261579,  c  =  .0608657,  A  =  115°  42'. 
10.  Given  a  =  35384.82,  c  =  57946.34,  B  =  19°  37'  25". 

116.   CASE  III.     Given  the  three  sides. 

The  angles  might  be  calculated  by  the  formulae  of  §  110 ;  but  as  these 
are  not  adapted  to  logarithmic  computation,  it  is  usually  more  convenient 
to  use  the  formulae  of  §  111. 

Each  of  the  three  angles  should  be  computed  trigonometrically,  for  we 
then  have  a  check  on  the  work,  since  their  sum  should  be  180°. 

If  all  the  angles  are  to  be  computed,  the  tangent  formulae  are  the  most 
convenient,  since  only  four  different  logarithms  are  required.  If  but  one 
angle  is  required,  the  cosine  formula  will  be  found  to  involve  the  least 
work. 

The  triangle  is  possible  for  any  values  of  the  data,  provided  no  side  is 
greater  than  the  sum  of  the  other  two. 

If  all  the  angles  are  required,  and  the  tangent  formulae  are  used,  it  is 
convenient  to  modify  them  as  follows ;  by  (65), 


l(s  —  a)(s  —  b}  (s  —  c)  , 

Denoting  \P  -  ^  -  &  -  '-  by  r,  we  have 
~  s 


tan  J  A  — 


r 


—  a 


In  like  manner,  tan  -i-  B  —  — - — .  and  tan  4-  C  =  — — • 

s  —  b  s  —  c 

1.   Given  a  =  2.51,  b  =  2.79,  c  =  2.33 ;  find  A,  B,  and  C. 
Here,  2s  =  a  +  &  +  c  =  7.63. 

Whence,  s  =  3.815,  s-a  =  1.305,  s  -  b  =  1.025,  s-c  =  1.485. 
We  have,  log  r  =  %  [log  (s  —  a)  +  log  (s  —  b)  +  log  (s  —  c)  +  colog  s}. 
Also,  log  tan  $A  —  log  r  —  log  (s  —  a), 

log  tan  %  B  =  log  r  —  log  (s  —  6), 
and  log  tan  £  C  =  log  r  —  log  (s  —  c> 


70 


PLANE  TRIGONOMETRY. 


log  (a- a)  =  0.115611 

log(s-6)  =  0.010724 

log  (s  -  c)  =  0.171726 

colog  s  =  9.418505-10 
2)19.716566  ^20 

logr  =  9.858283-10 

log  (8  -  a)  =  0.115611 

log  tan  1.4=  9.742672-10 


logr 

log  (s  -  6) 
log  tan  £  B 


logr  = 

log  (S  -  C)  : 

log  tan  ^(7 


.4  =  57°  52'  45.4". 

Check,  4.  +  B'+C=  179° 59' 59.6". 
2.   Given  a  =  7,  6  =  11,  c  =  9.6 ;  find  B. 


By  (63), 

Whence,  log  cos 
Here,  2s  = 


9.858283  - 10 
:  0.010724 
9.847559  - 10 
35°8'40.9". 
709 17' 21.8". 

9.858283  - 10 
0.171726 
9.686557  - 10 
25°  54' 56.2". 
51°  49' 52.4". 


=  -g-  [log  s  +  log  (*  —  6)  +  colog  c  +  colog  a]. 
+  c  =  27.6  ;  whence,  s  =  13.8,  and  s  —  6  =  2.8. 

logs=    1.139879 
log  0-5)=   0.447158 
colog  c=    9.017729-10 
colog  a  =   9.154902-10 
2)19.759668  -  20 
log  cos  £  3=   9.879834-10 

=  40°41'11.5",  and  B  =  81°  22  '23.0". 


EXAMPLES. 

Solve  the  following  triangles  : 

3.  Given  a  =  2, 

4.  Given  a  =  5, 

5.  Given  a  =  10, 

6.  Given  a  =  5.6, 

7.  Given  a  =  .85, 

8.  Given  a  =  61.3,    b  =  84.7,    c  =  47.6. 

9.  Given  a  =  705,     6  =  562,     c  =  639  ;  find  A. 

10.  Given  a  =  .0291,  6  =  .0184,  c  =  .0358  ;  find  B. 

11.  Given  a  =  3019,    6  =  6731,   c  =  4228j  find  C. 


6  =  3, 
6  =  7, 
6  =  9, 
6  =  4.3, 
6  =  :92, 

c  =  4. 
c  =  6. 
c  =  8. 
c  =  4.9. 
c  =  .78. 

SOLUTION   OF  OBLIQUE   TRIANGLES.  71 

117.   CASE  IY.     Given  two  sides  and  the  angle  opposite  to  one  of  them. 

It  was  stated  in  §  97  that  a  triangle  is,  in  general,  completely  deter- 
mined when  three  of  its  elements  are  known,  provided  one  of  them  is 
a  side.  The  only  exceptions  occur  in  Case  IY. 

To  illustrate,  let  us  consider  the  following  example : 

Given     a  =  52.1,  b  =  61.2,  A  =  31°  26' ;  find  JB,  O,  and  c. 

By  (47),  ™*  =  *,  or  sin  B  =  &sinA 

sin  A     a  a 

Whence,  log  sin  B  =  log  b  +  colog  a  +  log  sin  A. 

log  b  =  1.786751 
colog  a  =  8.283162  -  10 
log  sin  A  =  9.717259  - 10 

log  sin  B  =  9.787172  -  10 

B  =  37°  46' 37.9",  from  the  table. 

But  in  determining  the  angle  corresponding,  attention  must  be  paid 
to  the  fact  that  an  angle  and  its  supplement  have  the  same  sine  (§  33). 

Hence,  another  value  of  B  will  be  180° -37°  46 '37.9",  or  142°13'22.1"; 
and  calling  these  values  BI  and  B2,  we  have 

Bl  =  37° 46' 37.9",  and  B2  =  142°  13' 22.1". 

Note.  The  reason  for  this  ambiguity  is  at  once  apparent  when  we  attempt  to 
construct  the  triangle  from  the  data. 


OS? 


We  first  lay  off  the  angle  DAF  equal  to  31°  26',  and  on  AF  take  -4(7  =  61. 
With  C  as  a  centre,  and  a  radius  equal  to  52.1,  describe  an  arc  cutting  AD  at 
and  J52.     Then  either  of  the  triangles  AB\G  or  AB2C  satisfies  the  given  conditions. 

The  two  values  of  B  which  were  obtained  are  the  values  of  the  angles 
and  AB2C,  respectively;  and  it  is  evident  geometrically  that  these  angles  are  supple- 
mentary. 

To  complete  the  solution,  denote  the  angles  ACBl  and  ACB2  by  d 
and  C2,  and  the  sides  ABl  and  AB2  by  GI  and  c2. 

Then,  Ci  =  180°  -  (A  +  B,)  =  180°  -    69°  12' 37.9"  =  110° 47 '22.1", 
and  C2  =  180°  -  (-4  +  Bj)  =  180°  - 173°  39' 22.1"  =     6°  20' 37.9". 


72  PLANE  TRIGONOMETRY. 

Again,  by  («),         S  =  ™5  and  &-«*& 
a      sm  A  a      smA 

Whence,       c1  =  a  sin  Cj  esc  J.,  and  c2  =  a  sin  C2  esc  A 

log  a  =  1.716838  log  a  =  1.716838 

log  sin  Ci  =  9.970761  -  10  log  sin  02  =  9.043343  - 10 

log  esc  A  =  0.282741  log  esc  A  ==  0.282741 

log  G!  =  1.970340  log  Ca  =  1.042922 

d=  93.3985.  c2  =  11.0388. 

118.  Whenever  an  angle  of  an  oblique  triangle  is  determined  from  its 
sine,  both  the  acute  and  obtuse  values  must  be  retained  as   solutions, 
unless  one  of  them  can  be  shown  by  other  considerations  to  be  inadmissi- 
ble ;  and  hence  there  may  sometimes  be  two  solutions,  sometimes  one,  and 
sometimes  none,  in  an  example  under  Case  IV. 

1.  Let  the  data  be  a,  6,  and  A,  and  suppose  b  <  a. 

By  Geometry,  B  must  be  <  A ;  hence,  only  the  acute  value  of  B  can 
be  taken ;  in  this  case  there  is  but  one  solution. 

2.  Let  the  data  be  a,  b,  and  A,  and  suppose  b  >  a. 

Since  B  must  be  >  A,  the  triangle  is  impossible  unless  A  is  acute. 

Again,  since  — —  =  _  and  b  is  >  a,  sin  B  is  >  sin  A. 
sm  A      a 

Hence,  both  the  acute  and  obtuse  values  of  B  are  >  A,  and  there  are 
two  solutions,  except  in  the  following  cases : 

If  log  sin  B  =  0,  then  sin  B  =  1  (§  72),  and  B  =  90°,  and  the  triangle 
is  a  right  triangle ;  if  log  sin  B  is  positive,  then  sin  B  is  >  1,  and  the  tri- 
angle is  impossible. 

119.  The  results  of  §  118  may  be  stated  as  follows : 

If,  of  the  given  sides,  that  adjacent  to  the  given  angle  is  the  less,  there 
is  but  one  solution,  corresponding  to  the  acute  value  of  the  opposite  angle. 

If  the  side  adjacent  to  the  given  angle  is  the  greater,  there  are  two 
solutions,  unless  the  log  sine  of  the  opposite  angle  is  0  or  positive ;  in 
which  cases  there  are  one  solution  (a  right  triangle),  and  no  solution, 
respectively. 

120.  We  will  illustrate  the  above  points  by  examples : 
1.   Given  a  =  7.42,  6  =  3.39,  A  =  105°  13';  find  B. 

Since  6  is  <  a,  there  is  but  one  solution,  corresponding  to  the  acute 
value  of  B. 


SOLUTION  OF   OBLIQUE   TRIANGLES.  73 

By  (47),  sin  B  = 

Whence,  log  sin  B  =  log  b  +  colog  a  -f  log  sin  A. 

log  b  =  0.530200 
colog  a  =  9.129596  -10 
log  sin  A  =  9.984500  -10 

log  sin  B  =  9.644296  -10 
B  =  26°  9'  30.5". 

2.  Given  b  =  3,  c  =  2,  (7=100°;  find  B. 

Since  6  is  >  c,  and  C  is  obtuse,  the  triangle  is  impossible 

3.  Given  a  =  22.7643,  c  =  50,  .4  =  27°  5';  find  C. 

We  have,  sin  C  =  csinA. 

a 

log  c  =  1.698970 
colog  a  =  8.642746  -10 
log  sin  A  =  9.658284  -10 

log  sin  0=0.000000 

Therefore,  sin  (7=1,  and  (7  =  90°. 

Here  there  is  but  one  solution ;  a  right  triangle. 

4.  Given  a  =  .83,  b  =  .715,  B  =  61°  47';  find  A 

We  have,  sin^  =  £*5j?. 

6 

log  a  =  9.919078  -10 
colog  b  =  0.145694 
log  sin  B  =  9.945058  -  10 

log  sin  A  =  0.009830 
Since  log  sin  A  is  positive,  the  triangle  is  impossible. 

EXAMPLES. 

121.   Solve  the  following  triangles : 

1.   Given  a  =  5.98,          6  =  3.59,  .4  =  63°  50'. 

2    Given  b  =  74.1,          c  =  64.2,  (7  =  27°18P. 

3.  Given  b  =  .2337,         c=.0982,  5  =  108°. 

4.  Given  a  =  4.254,         c  =  4.536,  C  =  37°  9'. 

5.  Given  a  =  .2789,        b  =  .2271,  B  =  65°  38'. 


74  PLANE  TRIGONOMETRY. 

6.  Given  a  =  60.935,  c  =  76.097,  ^1  =  133°  41'. 

7.  Given  b  =  74.8067,  c  =  98.7385,  C  =  81°  47'. 

8.  Given  a  =  9.51987,  c  =  ll,  ^  =  59°  56'. 

9.  Given  b  =  4.521,  c  =  5.03,  B  =  40°  32'  7". 

10.  Given  a  =  186.82,  6  =  394.2,    B  =  114°  29'  51". 

11.  Given  6  =  5143.4,  c  =  4795.56,  C  =  72°  53' 38". 

12.  Given  a  =  .860619,  c=. 635761,  ^1  =  19°  12' 43". 
13. '  Given  a  =  139.27,  b  =  195.9716,  A  =  45°  17'  20". 

14.  Given  a  =.32163,   c=. 27083,   C=  52°  24' 16". 

15.  Given  b  =  91139.04,  c  =  80640.37,  B  =  126°  5 '  34". 

AREA  OP  AN  OBLIQUE  TRIANGLE. 
122.   1.   Given  a  =  18.063,  A  =  96°  30'  15",  B  =  35°  0'  13";  find  K 

By  (71),          2/f=a2sin^fn°=a2sin.BsinCfcscA 
sin  .4 

Whence,  log  (2  K)  —  2  log  a  +  log  sin  B  +  log  sin  C  +  log  esc  A 
From  the  data,    C  =  180°  -  (A  +  B)  =  48°  29'  32". 

log  a  =  1.256790  j  multiply  by  2  =  2.513580 

log  sin  5  =  9.758630 -10 
log  sin  C  =9.874404  -10 
log  esc  .4  =  0.002804 

log  (2  K)=  2.149418 

2  K=  141.065. 
^=70.533. 

EXAMPLES. 
Find  the  areas  of  the  following  triangles : 

2.  Given  a  =  38.09,          c  =  11.2,  JB  =  67°55'. 

3.  Given  a  =  5,  6  =  8,  c  =  6. 

4.  Given  b  =  6.074,        A  =  70°  39',  B  =  56°  23'. 

5.  Given  b  =  761.86,        c  =  526.02,  A  =  124°  6'  13". 

6.  Given  a  =  97,  b  =  83,  c  =  71. 

7.  Given  a  =  1.9375,      .4  =  43°  18',  B  =  29°  47'  36". 


SOLUTION  OF   OBLIQUE  TRIANGLES.  75 

8.  Given  b  =  .439592,  ^  =  62°  40' 8",   (7=  54°  32' 25". 

9.  Given  a  =  39.5,     b  =  44.8,       c  =  52.3. 

10.  Given  a  =  .804639,   c=. 357173,    B  =  18°  11'  49". 

11.  Given  c  =  95.86157,  B  =  115°  24'  52",  C  =  32°  57'  21". 

12.  Given  a  =  .02409481,  6  =  .02763834,   C  =  81°  9'  34". 

13.  Given  a  =  7.825,    b  =  6.592,      c  =  9.643. 

MISCELLANEOUS    EXAMPLES. 

123.  1.  From  a  point  in  the  same  horizontal  plane  with  the  base  of 
a  tower,  the  angle  of  elevation  of  its  top  is  52°  39',  and  from  a  point 
100  ft.  further  away  it  is  35°  16'.  Find  the  height  of  the  tower,  and  its 
distance  from  each  point  of  observation. 

2.  One  side  of  a  parallelogram  is  56,  and  the  angles  between  this  side 
and  the  diagonals  are  31°  14'  and  45°  37'.     Find  all  the  sides  of  the 
parallelogram. 

3.  In  a  field  ABCD,  the  sides  AB,  BC,  CD,  and  DA  are  155,  236, 
252,  and  105  rods,  respectively,  and  the  diagonal  AC  is  311  rods.     Find 
the  area  of  the  field. 

4.  The  area  of  a  triangle  is  1356,  and  two  of  its  sides  are  53  and  69. 
Find  the  angle  between  them. 

^  5.  From  the  top  of  a  bluff,  the  angles  of  depression  of  two  posts  in 
the  plain  below,  in  line  with  the  observer  and  1000  ft.  apart,  are  found  to 
be  27°  40'  and  9°  33',  respectively.  Find  the  height  of  the  bluff  above  the 
plain. 

6.  The  parallel  sides  of  a  trapezoid  are1 86  and  138,  and  the  angles  at 
the  extremities  of  the  latter  are  53°  49'  and  67°  55'.     Find  the  non-parallel 
sides. 

7.  Two  trains  start  at  the  same  time  from  the  same  point,  and  move 
along  straight  railways,  which  intersect  at  an  angle  of  74°  30',  at  the  rates 
of  30  and  45  miles  an  hour,  respectively.     How  far  apart  are  they  at  the 
end  of  45  minutes  ? 

8.  Two  sides  of  a  triangle  are  .5623  and  .4977,  and  the  difference  of 
the  angles  opposite  these  sides  is  15°  48'  32".     Solve  the  triangle. 

9.  Two  yachts  start  at  the  same  time  from  the  same  point,  and  sail, 
one  due  north  at  the  rate  of  10.44  mijes  an  hour,  and  the  other  due  north- 
east at  the  rate  of  7.71  miles  an  hour.     What  is  the  bearing  of  the  first 
yacht  from  the  second  at  the  end  of  half  an  hour  ? 


76  PLANE  TRIGONOMETRY. 

10.  A  vessel  is  sailing  due  south-west  at  the  rate  of  8  miles  an  hour. 
At  10.30  A.M.,  a  lighthouse  is  observed  to  bear  30°  west  of  north,  and  at 
12.15  P.M.,  it  is  observed  to  bear  15°  east  of  north.     Find  the  distance  of 
the  lighthouse  from  each  position  of  the  vessel. 

11.  Two  sides  of  a  parallelogram  are  65  and  133,  and  one  of  the 
diagonals  is  159.     Find  the  angles  of  the  parallelogram,  and  the  other 
diagonal. 

12.  To  find  the  distance  of  an  inaccessible  object  A  from  a  position  B, 
I  measure  a  line  BC,  208.3  ft.  in  length.     The  angles  ABC  and  ACB  are 
measured,  and  found  to  be  126°  35'  and  31°  48',  respectively.     Find  the 
distance  AB. 

13.  The  diagonals  of  a  parallelogram  are  81  and  106,  and  the  angle 
between  them  is  29°  18'.     Find  the  sides  and  angles  of  the  parallelogram. 

14.  A  flagpole  40  ft.  high  stands  on  the  top  of  a  tower.     From  a  posi- 
tion near  the  base  of  the  tower,  the  angles  of  elevation  of  the  top  and 
bottom  of  the  pole  are  38°  53'  and  20°  18',  respectively.     Find  the  distance 
and  height  of  the  tower. 

15.  AD  and  BC  are  the  parallel  sides  of  a  trapezoid  ABCD;  the  sides 
AB  and  BC  are  7.8  and  9.4,  respectively,  and  the  angles  B  and  C  are 
113°  47'  and  125°  34',  respectively.     Find  AD  and  CD. 

16.  A  surveyor  observes  that  his  position  A  is  exactly  in  line  with 
two  inaccessible  objects  B  and  C.     He  measures  a  line  AD  500  ft.  long, 
making  the  angle  BAD  =  60°,  and  at  D  observes  the  angles  ADB  and 
BDC  to  be  40°  and  60°,  respectively.     Find  the  distance  BC. 

17.  A  side  of  a  parallelogram  is  48,  a  diagonal  is  73,  and  the  angle 
between  the  diagonals,  opposite  the  given  side,  is  98°  6'.     Find  the  other 
diagonal  and  the  other  side. 

18.  To  find  the  distance  between  two  buoys  A  and  B,  I  measure  a 
base-line  CD  on  the  shore,  150  ft.  long.     At  the  point  C  the  angles  ACD 
and  BCD  are  measured,  and  found  to  be  95°  and  70°,  respectively ;  and 
at  D  the  angles  BDC  and  ADC  are  found  to  be  83°  and  30°,  respectively. 
Find  the  distance  between  the  buoys. 

19.  The  sides  AB,  BC,  and  CD,  of  a  quadrilateral  ABCD  are  38,  55, 
and  42,  respectively,  and  the  angles  B  and  C  are  132°  56'  and  98°  29', 
respectively.     Find  the  side  AD,  and  the  angles  A  and  D. 

20.  The  sides  AB,  BC,  and  DA  of  a  field  ABCD  are  37,  63,  and  20 

rods,  respectively,  and  the  diagonals  AC  and  BD  are  75  and  42  rods, 
respectively.     Find  the  area  of  the  field. 


CUBIC  EQUATIONS.  77 


IX.    CUBIC  EQUATIONS. 

124.  We  know,  by  Algebra,  that  a  cubic  equation  can  always  be 
transformed  into  another  in  which  the  term  containing  the  square  of  the 
unknown  quantity  shall  be  wanting. 

V) 

Thus,  if  the  equation  is  ic3  +ps?  +  qx  +  r  =  0,  putting  x  —  y  —  ^,  we 
have 


or,  y°- 

which  is  in  the  required  form. 

125.  Cardan's  Method  enables  us  to  solve  any  cubic  equation  of  the 

y-y3 

form  a?3  +  ax  +  b  =  0,  except  in  the  case  where  a  is  negative,  and  — 

b2 

numerically  > — • 
4 

In  this  case,  it  is  possible  to  find  the  roots  by  Trigonometry. 

126.  Trigonometric  Solution  of  Cubic  Equations. 

To  solve  the  equation 


a3      62 
where  a  is  positive,  and    ->- 


Putting  x  —  2  ra  cos  A,  the  equation  becomes 

m2  ~~  2  m1 


8  m3  cos3  A  —  2  am  cos  A  —  b  =  0,  or  4  cos3  A  —  —a  cos  A — -  =  0. 

/yy>2  Q  ,yv>3 


But  by  (36),  4  cos3  A  =  cos  3  A  +  3  cos  A. 

Whence,         cos  3  A  +  3  cos  A  --^-cos  A —  =  0. 

m2  2m3 

We  may  take  m  so  that  3  -  —2  =  0 ;  then,  3  m2  =  a,  and  m  =-v/-»     (B) 

Then  (A)  become!  cos  3  A  =  — « 

2  m* 


78  PLANE   TRIGONOMETRY. 

Substituting  in  this  the  value  of  m  from  (B),  we  have 


Since,  by  hypothesis,  —  <  ^->  "we  have  —  x  — £  <  !• 
4      27  4      a3 


Taking  the  square  root  of  each  member  of  the  inequality,  -A/-j  <  !• 
Hence,  the  value  of  3  A  in  (C)  is  possible,  since  its  cosin< 
Let  z  be  the  least  positive  angle  whose  cosine  is  equal  to 


Hence,  the  value  of  3  A  in  (C)  is  possible,  since  its  cosine  is  <  1. 

6/27 
2\a3' 

Then,  one  value  of  3  A  is  z;  and  all  its  values  are  given  by  the 
expression  2  UTT  ±  z  (§  62),  where  n  is  0  or  any  positive  or  negative 
integer. 

Whence,  cos  A  *=  cos  J  (2  nir  ±  z). 

Now  let  n  =  3  g  +  n',  where  q  is  0  or  any  positive  or  negative  integer, 
and  n'  =  0  or  ±  1  ;  then, 

cos  A  =  cos  (§!±2^£±^  =  cos  ("2  j,  +  ^±s]  =  cos  2»V±*  . 
o  o  o 

for  by  §  21,  any  multiple  of  360°  may  be  added  to,  or  subtracted  from, 
an  angle,  without  altering  its  functions. 
Putting  n1  =  0,  1,  and  —  1,  we  have 


=  c«/±!\ 

\      6J 


or  Cos=cos      or 


for  by  §  29,  the  cosine  of  the  angle  (—  A)  is  equal  to  the  cosine  of  A. 
But  x  =  2  m  cos  A  ;  and  hence  the  three  values  of  x  are 


cos|, 


where  z  is  given  by  the  equation  cos  2  =-— 


EXAMPLES. 

1.   Solve  the  equation  x3  —  4a;  +  2  =  0. 
Here  a  =  4,  6  =  -  2  ;  then,  cos  z  =  -  ^?|,  or  cos  (*-*)=          (§  33). 

By  logarithms,   log  cos  (TT  -  z)  =  £  (log  27  -  log  64). 

log  27=    1.431364 
log  64  =    1.806180 

2)19.625184-20 
log  cos  (TT  -  z)  =   9.812592  -  10 


UNIVERSITY 

OF 


CUBIC   EQUATIONS.  79 

Then,  TT  -  z  =  49°  29f  40.5",  and  2  =  130°  30'  19.5". 

Whence,          |  =  43°  30'  6.5",  and  2^|=  2^=  ^ 
Then  the  three  values  of  a;  are 


cos  (120°  -  43°  30'  6.6")  =\/cos  76°  29'  53.5", 
3  *  o 

and  -\Afcos  (120°  +  43°  30'  6.5")  =  A/—  cos  (90°  +  73°  30'  6.5") 

^  * 


Now, 


-  _^sin  73°  30'  6.5"  (§  30). 
3 


log         =  %  (log  16  -  log  3)  =  |  (1.204120  -  .477121)  =  .363500.       (1) 

*  o 

Also,        log  cos  43°  30'   6.5"  =  9.860549  -  10,  (2) 

log  cos  76°  29'  53.5"  =  9.368242  -  10,  (3) 

and  log  sin  73°  30'   6.5"  =  9.981741  -  10.  (4) 

Adding  (2),  (3),  and  (4)  in  succession  to  (1),  the  logarithms  of  the 
absolute  values  of  x  are 

0.224049,  9.731742  -  10,  and  0.345241. 
The  numbers  corresponding  to  these  logarithms  are 

1.67513,  .53919,  and  2.21432. 
Whence,  a  =  1.67513,  .53919,  or  -2.21432. 

Solve  the  following  equations  : 

2.  a?-4aj-l  =  0.  4.   ^  +  6^-0-1  =  0. 

3.  a^-6a;  +  3  =  0.  6.  arj-3arj-2a;  +  l  =0. 


SPHERICAL  TRIGONOMETRY. 


X.    GEOMETRICAL  PRINCIPLES. 

127.  If  a  triedral  angle  be  formed  with  its  vertex  at  the  centre  of  a 
sphere,  it  intercepts  on  the  surface  a  spherical  triangle. 

The  triangle  is  bounded  by  three  arcs  of  great  circles,  called  its  sides, 
which  measure  the  face  angles  of  the  triedral  angle. 

The  angles  of  the  spherical  triangle  are  the  spherical  angles  formed  by 
the  adjacent  sides ;  and,  by  Geometry,  each  is  equal  to  the  angle  between 
two  straight  lines  drawn,  one  in  the  plane  of  each  of  its  sides,  and  per- 
pendicular to  the  intersection  of  these  planes  at  the  same  point. 

128.  The  sides  of  a  spherical  triangle  are  usually  expressed  in  degrees. 

129.  Spherical    Trigonometry   treats    of   the    trigonometric   relations 
between  the  sides  and  angles  of  a  spherical  triangle. 

The  face  and  diedral  angles  of  the  triedral  angle  are  not  altered  by 
varying  the  radius  of  the  sphere;  and  hence  the  relations  between  the 
sides  and  angles  of  a  spherical  triangle  are  independent  of  the  length  of 
the  radius. 

130.  We  shall  limit  ourselves  in  the  present  work  to  such  triangles 
as  are  considered  in  Geometry,  where  each  angle  is  less  than  two  right 
angles,  and  each  side  less  than  the  semi-circumference  of  a  great  circle*, 
that  is,  where  each  element  is  less  than  180°. 

131.  The  proofs  of  the  following  properties  of  spherical  triangles  may 
be  found  in  any  treatise  on  Solid  Geometry : 

1.  The  sum  of  any  two  sides  of  a  spherical  triangle  is  greater  than 
the  third  side. 

2.  In  any  spherical  triangle,  the  greater  side  lies  opposite  the  greater 
angle ;  and,  conversely,  the  greater  angle  lies  opposite  the  greater  side. 

3.  The  sum  of  the  sides  of  a  spherical  triangle  is  less  than  360°. 

80 


GEOMETRICAL  PRINCIPLES.  81 

4.  The  sum  of  the  angles  of  a  spherical  triangle  is  greater  than  180°, 
and  less  than  540°. 

5.  If  A'B'C'  is  the  polar  triangle  of  ABC,  that  is,  if  A,  B,  and  0 
are  the  poles  of  the  sides  B'C',  C'A',  and  A'B'j  respectively,  then,  con- 
versely, ABC  is  the  polar  triangle  of  A'B'G1. 


6.  In  two  polar  triangles,  each  angle  of  one  is  measured  by  the  sup- 
plement of  the  side  lying  opposite  the  homologous  angle  of  the  other; 
that  is 

a'  =180°  -A.  b'  =  lSQ°-B.  c'  =  180°-a 


^'  =  180°  -a.  B'  =  lSQ°-b.  C'  =  180°-c. 

132.  A  spherical  triangle  is  called  tri-rectangular  when  it  has  three 
right  angles  ;  each  side  is  a  quadrant,  and  each  vertex  is  the  pole  of  the 
opposite  side. 

133.  I.   Let  C  be  the  right  angle  of  the  right  spherical  triangle  ABC, 
and  suppose  a  <  90°  and  b  <  90°. 


Complete  the  tri-rectangular  triangle  A'B'C;  also,  since  B'  is  the  pole 
of  AC}  and  A'  of  BC,  construct  the  tri-rectangular  triangles  AB'D  and 
ABE. 

Then  since  B  lies  within  the  triangle  AB'D,  AB  or  c  is  <  90°. 

Since  BC  is  <  B'C}  the  angle  A  is  <  B'AD,  or  <  90°. 

Since  AC  is  <  A'C,  the  angle  B  is  <  A'BE,  or  <  90°. 


82  SPHERICAL  TRIGONOMETRY. 

II.   Suppose  a  <  90°  and  b  >90°. 

c JB 


Complete  the  lune  ABA'C. 

Then  in  the  right  triangle  ABC,  A'C=1SO°  -  b. 
That  is,  the  sides  a  and  AC  of  the  triangle  ABC  are  each  <  90° ;  and 
by  L,  A'B  and  the  angles  A  and  ABC  are  each  <  90°. 
But,  c  =  180°  -  AB,  A  =  A,  and  5  =  180°  -  ABC. 
Whence,  c  is  >  90°,  A  <  90°,  and  B  >  90°. 
Similarly,  if  a  is  >  90°  and  b  <  90°,  then  c  is  >  90°,  A  >  90°,  and 


III.   Suppose  a  >  90°  and  b  >90°. 

a 


b 

Complete  the  lune  ACBC'. 

Then  in  the  right  triangle  ABC',  AC'  =  180°  -  b,  and  BC'  =  180°  -  a. 
That  is,  the  sides  AC'  and  BC'  of  the  triangle  ABC'  are  each  <  90° ; 
and  by  I.,  AB  and  the  angles  BAG'  and  ABC'  are  each  <  90°. 
But,  A  =  180°  -  BAG',  and  B  =  180°  -  ABC'. 
Whence,  c  is  <  90°,  A  >  90°,  and  B  >  90°. 
Hence,  in  any  right  spherical  triangle : 

1.  If  the  sides  about  the  right  angle  are  in  the  same  quadrant,  the  hypote- 
nuse is  <  90° ;  if  they  are  in  different  quadrants,  the  hypotenuse  is  >  90°. 

2.  An  angle  is  in  the  same  quadrant  as  its  opposite  side. 

134.  In  the  figure  of  §  131,  we  have,  by  §  131, 1,  a'  <  b'  -f  c'. 
Putting  for  a',  b',  and  c'  the  values  given  in  §  131,  6,  we  have 

180°  -  A  <  180°  -  B  -f  180°  -  C,  or  B  +  C  -  A<  180°. 

Again,  by  §  130,  B  +  C  +  180°  >  A ;  whence,  B  +  C  -  A  >  - 180\ 

Therefore,  B  +  C-  A  is  between  180°  and  -  180°. 

Similarly,  C  +  A-B  and  A  +  B  -  C  are  between  180°  and  -  180°. 


RIGHT  SPHERICAL   TRIANGLES. 

\ 

XI.    RIGHT  SPHERICAL  TRIANGLES. 

135.   Let  C  be  the  right  angle  of  the  right  spherical  triangle  ABC. 

B 


83 


Let  0  be  the  centre  of  the  sphere,  and  draw  OA,  OB,  and  0(7. 

At  any  point  A'  of  OA  draw  A'B'  and  A'C'  perpendicular  to  OA,  meet- 
ing OB  and  0(7  at  B'  and  C',  and  draw  £'0'. 

Then  OA  is  perpendicular  to  the  plane  A  B'C'. 

Hence,  each  of  the  planes  A'B'C'  and  OBC  is  perpendicular  to  the 
plane  OAC,  and  their  intersection  B'C'  is  perpendicular  to  0-4(7. 

Therefore,  B'C'  is  perpendicular  to  A'C'  and  00'. 

In  the  right  triangle  OA'B',  we  have 

OA' 
OC1' 


lA'OB'-OA'-00' 
~~-~ 


But  in  the  right  triangles  OB'C'  and  OC'A', 


Whence, 
Again, 


cos  c  =  cos  a  cos  b. 
B'C' 


B'C1      OB'  _  sina 

^B'~.4'.B'~smc' 

OB' 

A'C' 


A       ,3  -DIAim         A'C'  OA'          tail& 

And,  cos  A  =  cos  B'A'C'  =  -77^7  =  -77^  =  z 

7  alHlArUI          4-n-n  /» 


JL'5'     ^l'J3y     tanc 
OA' 


In  like  manner, 


sin  3  =          , 

sine 


and 


cos  B 


tano^ 
tanc 


(75) 
(76) 

(77) 

(78) 
(79) 


84  SPHERICAL  TRIGONOMETRY. 

136.  The  proofs  of  §  135  cannot  be  regarded  as  general,  for  in  the 
construction  of  the  figure  we  have  assumed  a  and  b,  and  therefore  c  and 
A  (§  133),  to  be  less  than  90°. 

To  prove  formulae  (75)  to  (79)  universally,  we  must  consider  two  addi- 
tional cases : 

CASE  I.     When  one  of  the  sides  a  and  b  is  <  90°,  and  the  other  >  90°. 

c B 


In  the  right  spherical  triangle  ABC,  let  a  be  <  90°  and  b  >  90°. 
Complete  the  lune  ABAC',  then,  in  the  spherical  triangle  A'BC, 

A'B  =  180°  -  c,  AC  =  180°  -b,  A  =  A,  and  ABC  =  180°  -  B. 

But  by  §  133,  c  is  >  90°,  A  <  90°,  and  B  >  90°. 

Hence,  each  element,  except  the  right  angle,  of  the  right  spherical 
triangle  ABC  is  <  90° ;  and  we  have  by  §  135, 

cos  AB  =  cos  a  cos  AC, 


COS  .o.  — •    ,  ~— ~ — -  —  •"  A  i -rt 

taaAB  t&nA'B 

Putting  for  AB,  AC,  A  and  ABC  their  values,  we  have 
cos  (180°  -  c)  =  cos  a  cos  (180°  -  b), 
sin  a  .    ~  ™0  _  Rx  _  sin  (180° -6) 


cos(180°-£): 


•tan(180°-c)'  tan(180°-c) 

Whence,  by  §  33,         —  cos  c  =  cos  a  (—  cos  b), 


sine  •  sine 


^  T> 

—cos  B  = 


—  tan  c 


and  we  obtain  formulae  (75)  to  (79)  as  before. 

In  like  inanner,  the  formulas  may  be  proved  to  hold  when  a  is  >  90° 
and  b  <  90°. 


RIGHT   SPHERICAL  TRIANGLES.  85 

CASE  II.     When  both  a  and  b  are  >  90°. 


In  the  right  spherical  triangle  ABC)  let  a  and  b  be  >  90°. 
Complete  the  lune  ACBC'. 
By  §  133,  c  is  <  90°,  A  >  90°,  and  B  >  90°. 

Hence,  each  element,  except  the  right  angle,  of  the  right  spherical 
triangle  ABC'  is  <  90°  ;  and  we  have  by  §  135, 

cos  c  =  cos  AC1  cos  BCf, 


sm  c  sin  c 


, 

tan  c  tan  c 

Putting  for  AC',  BC',  BAG1,  and  ABC1  their  values,  we  have 
cos  c  =  cos  (180°  -  a)  cos  (180°  -  6), 


sin  c  sin  c 


tanc  tanc 

Whence,  by  §  33,       cos  c  =  (—  cos  a)  (—  cos  6), 


sin  c  sn  c 


tan  c  tan  c 

and  we  obtain  formulae  (75)  to  (79),  as  before. 

137.   From  (76)  and  (77),  we  obtain 

tan^L  =  sin  A  _  sing     tanc  _      sing 

cos  .4      sine     tan  b     cose  tan  6 


Whence  by  (75),        tan^L  = sn"         =  ^f .  (80) 

cos  a  cos  6  tan  b 


In  like  manner,          tan  B  =  HE£.  (81) 

sin- a  v    ' 


86  SPHERICAL  TRIGONOMETRY. 

138.   By  (4),  sin  a  =  cos  a  tan  a  ;  then  (76)  may  be  written 

tang 
cos  a  tan  a      tanc 


cos  c  tan  c      cose 
cos  a 
Whence  by  (75)  and  (79), 

£^.  (82) 

cos  6 


In  like  manner.          sin  B  =  .  (83) 

cos  a 

139.  From  (75),  (82),  and  (83),  we  have 


=  cosacos6  =  x  =  cot^lcot£.  (84) 

sin  .4 


140.  The  formulae  of  §§  135  to  139  are  collected  below  for  convenience 
jf  reference  : 

cos  c  =  cos  a  cos  b. 


sin  c  sin  c 


-.  cos     = 

tan  c  tanc 


sin  b  sin  a 


cos  o  cos  a 

cos  c  =  cot  -4  cot  B. 

The  student  should  compare  the  formulae  for  the  sines,  cosines,  and 
tangents  of  A  and  B  with  the  corresponding  formulae  in  §§  2  and  5. 

141.  Napier's  Knles  of  Circular  Parts. 

These  are  two  rules  which  include  all  the  formulae  of  §  140. 

co.  B 


co.  A 


In  any  right  spherical  triangle,  the  elements  a  and  6,  and  the  comple- 
ments of  the  elements  A,  B,  and  c  (written  in  abbreviated  form,  co.  A, 
co.  B,  and  co.  c),  are  called  the  circular  parts. 


RIGHT   SPHERICAL  TRIANGLES.  87 

If  we  suppose  them  arranged  in  the  order  in  which  the  letters  occur  in 
the  triangle,  any  one  of  the  five  may  be  taken  and  called  the  middle  part ; 
the  two  immediately  adjacent  are  called  the  adjacent  parts,  and  the 
remaining  two  the  opposite  parts. 

Then  Napier's  rules  are : 

I.    The  sine  of  the  middle  part  is  equal  to  the  product  of  the  tangents  of 
the  adjacent  parts. 

II.  The  sine  of  the  middle  part  is  equal  to  the  product  of  the  cosines  of 
the  opposite  parts. 

142.  Napier's  rules  may  be  proved  by  taking  each  circular  part  in 
succession  as  the  middle  part,  and  showing  that  the  results  agree  with  the 
formulae  of  §  140. 

1.  If  a  be  taken  as  the  middle  part,  b  and  co.  B  are  the  adjacent  parts, 
and  co.  c  and  co.  A  the  opposite  parts. 

Then  the  rules  give 

sin  a  =  tan  b  tan  (co.  jB),  and  sin  a  =  cos  (co.  c)  cos  (co.  A). 
Or  by  §  32,  sin  a  =  tan  b  cot  B,  and  sin  a  =  sin  c  sin  A ; 
which  are  equivalent  to  (81)  and  (76). 

2.  If  b  be  taken  as  the  middle  part,  a  and  co.  A  are  the  adjacent  parts, 
and  co.  c  and  co.  B  the  opposite  parts. 

Then,  sin  b  =  tan  a  tan  (co.  A)  =  tan  a  cot  A, 

and  sin  b  =  cos  (co.  c)  cos  (co.  B)  =  sin  c  sin  B\ 

which  are  equivalent  to  (80)  and  (78). 

3.  If  co.  c  be  taken  as  the  middle  part,  co.  A  and  co.  B  are  the  adjacent 
parts,  and  a  and  b  the  opposite  parts. 

Then,  sin  (co.  c)  =  tan  (co.  A)  tan  (co.  J3),  and  sin  (co.  c)  =  cos  a  cos  b. 
Or,  cos  c  =  cot  A  cot  5,  and  cos  c  =  cos  a  cos  b ; 

which  agree  with  (84)  and  (75). 

4.  If  co.  A  be  taken  as  the  middle  part,  b  and  co.  c  are  the  adjacent 
parts,  and  a  and  co.  B  the  opposite  parts. 

Then,  sin  (co.  A)=  tan  b  tan  (co.  c),  and  sin  (co.  A)= cos  a  cos  (co.  B). 
Or,  cos  A  =  tan  b  cot  c,  and  cos  A  =  cos  a  sin  B ; 

which  are  equivalent  to  (77)  and  (83). 


88  SPHERICAL  TRIGONOMETRY. 

5.  If  co.  B  be  taken  as  the  middle  part,  a  and  co.  c  are  the  adjacent 
parts,  and  b  and  co.  A  the  opposite  parts. 

Then,  sin  (co.  JB)=  tan  a  tan  (co.  c),  and  sin  (co.  B)=  cos  6  cos  (co.  A). 

Or,     \         cos  B  =  tan  a  cot  c,  and  cos  B  =  cos  b  sin  A ; 
which  are  equivalent  to  (79)  and  (82). 

Writers  on  Trigonometry  differ  as  to  the  practical  value  of  Napier's 
rules;  but  in  the  opinion  of  the  highest  authorities,  it  seems  to  be  re- 
garded as  preferable  to  attempt  to  remember  the  formulae  by  comparing 
them  with  the  analogous  formulae  for  plane  right  triangles,  as  stated  in 
§140. 

SOLUTION  OF  RIGHT  SPHERICAL  TRIANGLES. 

143.  To  solve  a  right  spherical  triangle,  two  elements  must  be  given 
in  addition  to  the  right  angle. 

There  may  be  six  cases : 

1.  Given  the  hypotenuse  and  an  adjacent  angle. 

2.  Given  an  angle  afod  its  opposite  side. 

3.  Given  an  angle  and  its  adjacent  side. 

4.  Given  the  hypotenuse  and  another  side. 

5.  Given  the  two  sides  a  and  b. 

6.  Given  the  two  angles  A  and  B. 

144.  Either  of  the  above  cases  may  be  solved  by  aid  of  §  140. 

The  formula  for  computing  either  of  the  remaining  elements  when  any 
two  are  given  may  be  found  by  the  following  rule : 

Take  that  formula  which  involves  the  given  parts  and  the  required  part. 

If  all  the  remaining  elements  are  required,  the  following  rule  may  be 
found  convenient  in  selecting  the  formulae : 

Take  the  three  formulae,  ivhich  involve  the  given  parts. 

145.  It  is  convenient  in  the  solution  to  have  a  check  on  the  log- 
arithmic work,  which  may  be  done  in  every  case  without  the  necessity  of 
looking  out  any  new  logarithms. 

Examples  of  this  will  be  found  in  §  148. 

The  check  formula  for  any  particular  case  may  be  selected  from  the 
set  in  §  140  by  the  following  rule : 

Take  that  formula  which  involves  the  three  required  parts. 

Note.  If  Napier's  rules  are  used,  the  following  rule  will  indicate  which  of  th« 
circular  parts  corresponding  to  the  given  elements  and  any  required  element  is  to  be 
regarded  as  the  middle  part. 


RIGHT  SPHERICAL  TRIANGLES.  89 

If  these  three  circular  parts  are  adjacent,  take  the  middle  one  as  the  middle  part, 
and  the  others  are  then  adjacent  parts. 

If  they  are  not  adjacent,  take  the  part  which  is  not  adjacent  to  either  of  the  others 
as  the  middle  part,  and  the  others  are  then  opposite  parts. 

For  the  check  formula,  proceed  as  above  with  the  circular  parts  corresponding  to 
the  three  required  elements. 

Thus,  if  c  and  A  are  the  given  elements, 

1.  To  find  a,  consider  the  circular  parts  a,  co.  c,  and  co.  A ;   of  these,  a  is  the 
middle  part,  and  co.  c  and  co.  A  are  opposite  parts.    Then,  by  Napier's  rules, 

sin  a  —  cos  (co.  c)  cos  (co.  A)  =  sin  c  sin  A. 

2.  To  find  b,  the  circular  parts  are  b,  co.  c,  and  co.  A ;  in  this  case  co.  A  is  the 
middle  part,  and  b  and  co.  c  are  adjacent  parts.    Then, 

sin  (co.  A)  =  tan  b  tan  (co.  c),  or  cos  A  —  tan  b  cot  c. 

3.  To  find  J5,  the  circular  parts  are  co.  B,  co.  c,  and  co.  A ;  co.  c  is  the  middle  part, 
and  co.  A  and  co.  B  are  adjacent  parts.    Then, 

sin  (co.  c)  =  tan  (co.  A)  tan  (co.  1?),  or  cos  c  =  cot  A  cot  B. 

4.  For  the  check  formula,  the  circular  parts  are  a,  6,  and  co.  B ;  a  is  the  middle 
part,  and  b  and  co.  B  are  adjacent  parts.    Then, 

sin  a  =  tan  b  tan  (co.  B)  =  tan  b  cot  B. 

146.  In  solving  spherical  triangles,  careful  attention  must  be  given  to 
the  algebraic  signs  of  the  functions ;  the  cosines,  tangents,  and  cotangents 
of  angles  between  90°  and  180°  being  taken  negative  (§  20). 

It  is  convenient  to  place  the  sign  of  each  function  just  above  or  below 
it,  as  shown  in  the  examples  of  §  148 ;  the  sign  of  the  function  in  the  first 
member  being  then  determined  in  accordance  with  the  principle  that  like 
signs  produce  +,  and  unlike  signs  produce  — . 

Note.  In  the  examples  after  the  first  of  §  148,  the  signs  are  omitted  in  every  case 
where  both  factors  of  the  second  member  are  + . 

147.  In  finding  the  angles  corresponding,  if  the  function  is  a  cosine, 
tangent,  or  cotangent,  its  sign  determines  whether  the  angle  is  acute  or 
obtuse ;  that  is,  if  it  is  +,  the  angle  is  acute ;  and  if  it  is  — ,  the  angle  is 
obtuse,  and  the  supplement  of  the  acute  angle  obtained  from  the  tables 
must  be  taken  (§  33). 

If  the  function  is  a  sine,  since  the  sine  of  an  angle  is  equal  to  the  sine 
of  its  supplement  (§  33),  both  the  acute  angle  obtained  from  the  tables 
and  its  supplement  must  be  retained  as  solutions,  unless  the  ambiguity 
can  be  removed  by  the  principles  of  §  133. 


90  SPHERICAL  TRIGONOMETRY. 

EXAMPLES. 

14a  1.   Given  B  =  33°  50',  a  =  108°  ;  find  A,  b,  and  c. 
By  the  rule  of  §  144,  the  formulae  from  §  140  are, 


=         ,  and 


cos  a       sin  a          tan  c 

—     —   +    +    +    -f       ~- 
Or,  cos  A  =  cos  a  sin  5,  tan  b  =  sin  a  tan  B,  and  tanc  = 


+ 
Hence,  log  cos  A  =  log  cos  a  +  log  sin  B. 

log  tan  b  =  log  sin  a  +  log  tan  5. 
log  tan  c  =  log  tan  a  —  log  cos  B. 

Since  cos  ^.  and  tan  c  are  negative,  the  supplements  of  the  acute  angles 
obtained  from  the  tables  must  be  taken  (§  147). 

Note  1.  When  the  supplement  of  the  angle  obtained  from  the  tables  is  to  be 
taken,  it  is  convenient  to  write  180°  minus  the  element  in  the  first  member,  as  shown 
below  in  the  cases  of  A  and  c. 

By  the  rule  of  §  145,  the  check  formula  for  this  case  is 

cos  A  —  JIEL^  or  log  cos  A  =  log  tan  b  —  log  tan  c. 
tan  c 

The  values  of  log  tan  b  and  log  tan  c  may  be  taken  from  the  first  part 
of  the  work,  and  their  difference  should  be  equal  to  the  result  previously 
found  for  log  cos  A. 

log  cos  a  =  9.489982  -  10  log  tan  a  =  0.488224 

log  sin  B  =  9.745683  -  10  log  cos  B  =  9.919424  -  10 

log  cos  A  =  9.235665  -  10  log  tan  c  =  0.568800 

180°  -A  =  80°  5'  33.8".  180°-c=   74°  53'  45.0". 

A  =  99°  54'  26.2".  c  =  105°  6'  15.0". 

log  sin  a  =  9.978206  -  10  Check. 

log  tan  B  =  9.826259  -  10  log  tan  b  =  9.804465  -  10 

logtanfc  =9.804465-10  loStanc  =  °'568800 

b  =  32°  30'  59.8".  log  cos^.  =  9.235665  -  10 

2.   Given  c  =  70°  30',  A  =  100°  ;  find  a,  6,  and  B. 
In  this  case  the  three  formulae  are 


—  ,  and  cos  c  =  cot  A  cot  B. 

sn  c  tan  c 

+  +       - 

Or,   sin  a  =  sin  c  sin  A,  tan  b  =  tan  c  cos  A,  and  cot  B  =  cos  c  tan  A. 


RIGHT  SPHERICAL  TRIANGLES.  91 

Here  the  side  a  is  determined  from  its  sine;  but  the  ambiguity  is 
removed  by  the  principles  of  §  133 ;  for  a  and  A  must  be  in  the  same 
quadrant.  Therefore,  a  is  obtuse;  and  the  supplement  of  the  angle 
obtained  from  the  tables  must  be  taken. 

By  §  145,  the  check  formula  is 

tan  B  =  -T— ,  or  sin  a  =  tan  b  cot  B. 

sin  a 

Note  2.  The  check  formula  should  always  be  expressed  in  terms  of  the  functions 
used  in  determining  the  required  parts ;  thus,  in  the  case  above,  the  check  formula  is 
transformed  so  as  to  involve  cot  B  instead  of  tan  B. 

log  sine   =9.974347-10  log  cose   =9.523495-10 

log  sin  A  =  9.993351  - 10  log  tan  J.  =  0.753681 

log  sin  a  =  9.967698  -  10  log  cot  B  =  0.277176 

180°  -  a  =  68°  10 '28.2".  180°  -  B  =  27°  50 '39.8". 

a  =  111°  49' 31.8".  B  =  152°  9' 20.2". 

log  tan  c  =0.450851 

log  cos  A  =  9.239670  -  10  Check. 

log  tan  b  =  9.690521  -  10  log  tan  b  =  9.690521  — 10 

180°  -  b  =  26°  7'  18.4".  loS  cot  B  =  °-27717^ 

b  =  153°  52'  41.6".  log  sin  a  =  9.967697  -  10 

Note  3.  We  observe  here  a  difference  of  .000001  in  the  two  values  of  log  sin  a. 
This  does  not  necessarily  indicate  an  error  in  the  work,  for  such  a  small  difference 
might  easily  be  due  to  the  fact  that  the  logarithms  are  only  approximately  correct  to 
the  sixth  decimal  place. 

3.   Given  a  =  132°  6',  b  =  77°  51' ;  find  A,  B}  and  c. 
In  this  case  the  three  formulae  are 


— ,  uc*-  ^  =  if^   and  cos  c  =  cos  a  cos  b. 
sm6  sma 

The  check  formula  is 

cos  c  =  cot  A  cot  B,  or  cos  c  tan  A  tan  5  =  1. 
That  is,       log  cos  c  4-  log  tan  A  +  log  tan  B  =  log  1  =  0. 

log  tan  a  =  0.044039  log  cos  a  =  9.826351  - 10 

log  sin  b    =9.990161-10  log  cos  b  =  9.323194  - 10 

log  tan  A  =  0.053878  log  cos  c  =  9.149545  - 10 

180°  -  A  =  48°  32'  41.8".  180°  -  c  =  81°  53'  17.4". 

A  =  131°  27'  18.2".  c  =  98°  6'  42.6". 


92  SPHERICAL  TRIGONOMETRY. 

Check. 

log  tan  &  =0.666967  log  cose   =9.149545-10 

log  sin  a  =9.870390-10  log  tan  A  =  0.053878 

log  tan  B  =0.796577  log  tan  £  =  0.796577 

B  =  80°  55'  26.6".  log  1  =  0.000000 

4.   Given  A  =  105°  59',  a  =  128°  33';  find  6,  B,  and  c. 
The  formulae  are 

•+ ,  _  tana      ^ -&  _  cos  A        d    in    =  s*nat 
tan^L'  *  cos  a'  sin .4 

The  check  formula  is  sin.B=^5— 

sin  c 

In  this  case,  each  of  the  required  parts  is  determined  from  its  sine ; 
and  as  the  ambiguity  cannot  be  removed  by  §  133,  both  the  acute  angle 
obtained  from  the  tables  and  its  supplement  must  be  retained  in  each 
case. 

log  tan  a  =0.098617  log  sin  a  =9.893243-10 

log  tan  A  =  0.542981  log  sin  A  =  9.982878  -  10 

log  sin 6   =  9.555636  - 10  log  sine    =9.910365-10 

6  =  21°  3'  58.7"  e  =  54°  26'  26.7" 

or  158°  56'  1.3".  or  125°  33'  33.3". 

Check. 

log  cos  A  =  9.439897  - 10  log  sin  b   =  9.555636  -  10 

log  cos  a  =9.794626-10  log  sine    =9.910365-10 

log  sin  B  =  9.645271  - 10  log  sin  B  =  9.645271  - 10 

£  =  26°  13'  18.2" 
or  153°  46' 41.8". 

It  does  not  follow,  however,  that  these  values  can  be  combined  pro- 
miscuously ;  for  by  §  133,  since  a  is  >  90°,  with  the  value  of  b  less  than 
90°  must  be  taken  the  value  of  c  greater  than  90°,  and  the  value  of  B  less 
than  90° ;  while  with  the  value  of  b  greater  than  90°  must  be  taken  the 
value  of  c  less  than  90°,  and  the  value  of  B  greater  than  90°. 

Thus  the  only  solutions  of  the  example  are : 

1.  b  =  21°  3'  58.7",  c  =  125°  33' 33.3",  B  =  26°  13 '18.2". 

2.  6  =  158°  56' 1.3",  c  =  54°  26 '26.7",  B  =  153°  46 '41.8". 


RIGHT  SPHERICAL  TRIANGLES. 


93 


Note  4.    The  figure  shows  geometrically  why  there  are  two  solutions  in  this  case. 

B 


For  if  AB  and  AC  be  produced  to  A\  forming  the  lune  ABA'C,  the  triangle 
A'BC  has  the  side  a  and  the  angle  A1  equal,  respectively,  to  the  side  a  and  the  angle 
A  of  the  triangle  ABC-,  and  both  triangles  are  right-angled  at  G. 

It  is  evident  that  the  sides  A'B  and  A'C  and  the  angle  A'BC  are  the  supplements 
of  the  sides  c  and  b  and  the  angle  ABC,  respectively. 

Solve  the  following  right  spherical  triangles : 


5.  Given   c  =  49°, 

6.  Given  ^  =  38°, 

7.  Given  ^  =  31°, 

8.  Given  B  =  153°, 

9.  Given  a  =  15°, 

10.  Given   c  =  139°, 

11.  Given  B  =  82°  25', 

12.  Given   c  =  75°37', 

13.  Given   c  =  118°49', 

14.  Given  a  =  171°  6', 

15.  Given  B  =  100°  40', 

16.  Given  A  =  102°  57', 

17.  Given  a  =  10°  28', 

18.  Given  A  =  54°  11', 

19.  Given  .4  =  50°  43', 

20.  Given   c  =  59°3', 

21.  Given  B  =  103°  30', 

22.  Given  A=  95°  15', 

23.  Given  c  =  78°52', 

24.  Given    c  =  127°9', 

25.  Given  A  =  98°  34', 

26.  Given   c  =  136°21', 


a  =  27°. 
B  =  63°. 
a  =  23°. 
a  =  35°. 
b  =  106°. 
J.  =  165°. 
b  =  68°  35'. 
J5  =  29°  4'. 
6  =  44°  23'. 
6  =  161°  58'. 
a  =  170°  38'. 
B  =  143°  46'. 
6  =  7°  10'.  - 
6  =  83°  29'. 
B  =  122°  18'. 
A  =  147°  32'. 
b  =  132°  54'. 
6  =  166°  7'. 
a  =  114°  26'. 
J5  =  80°51'. 
a  =113°  12'. 
6  =  157°  41'. 


94  SPHERICAL   TRIGONOMETRY. 

149.   Quadrantal  Triangles. 

A  spherical  triangle  is  called  quadrantal  when  it  has  one  side  equal  to 
a  quadrant. 

By  §  131,  6,  the  polar  triangle  of  a  quadrantal  triangle  is  a  right 
spherical  triangle. 

Hence,  to  solve  a  quadrantal  triangle,  we  have  only  to  solve  its  polar 
triangle,  and  take  the  supplements  of  the  results. 

1.   Given  c  =  90°,  a  =  67°  38',  b  =  48°  50' ;  find  A,  B,  and  C. 
Denoting  the  polar  triangle  by  A'B'C',  we  have  by  §  131,  6 : 

C'  =  90°,  A'  =  112°  22',  B'  =  131°  10' ;  to  find  a',  &',  and  c'. 
By  §  144,  the  formulae  for  the  solution  are 

cos  a'  =  52§4-'  cos  b'  =  ^ri>  and  cos  c'  =  co^'  coi"B'- 
sin  B'  sin  <AF 

+  -f 

The  check  formula  is  cos  c'  =  cos  a'  cos  b'. 

log  cos  A1  =  9.580392  -  10  log  cot  A1  =  9.614359  -  10 

log  sin  B1  =  9.876678  -  10  log  cot  B'  =  9.941713  -  10 

log  cos  a'  =9.703714-10  log  cose'   =9.556072-10 

180°  -  a'  =  59°  38'  9.7".  c'  =  68°  54'  41.5". 

log  cos  B'  =  9.818392  -  10  Check. 

log  sin  A'  =  9.966033  -  10  log  cos  a'  =  9.703714  -  10 

log  cos  b'  =  9.852359  -  10  log  cos  6'  =  9.852359  -  10 

180°  -  6'  =  44°  37'  5.8".  log  cos  c'  =  9.556073  -  10 

Then  in  the  given  quadrantal  triangle,  we  have 

A  =  180°  -  a'  =  59°  38' 9.7", 

B  =  180°  -  b'  =  44° 37' 5.8", 

O  =  180° -c'  =  111°  5' 18.5". 

EXAMPLES, 

Solve  the  following  quadrantal  triangles : 

2.  Given  A  =  122°,          6  =  154°. 

3.  Given  A=   45°  52',  B  =  139°  24'. 

4.  Given    a=    30°  19',   C=    42°  31'. 

5.  Given  B  =  51°  35',  C  =  116°  13'. 

6.  Given  A  =  105°  8',  a  =  104°  56r. 

7.  Given  a=  67°  27',  6=  81°  40'. 


RIGHT  SPHERICAL   TRIANGLES.  95 

150.  Isosceles  Spherical  Triangles. 

We  know,  by  Geometry,  that  if  an  arc  of  a  great  circle  be  drawn  from 
the  vertex  of  an  isosceles  spherical  triangle  to  the  middle  point  of  the 
base,  it  is  perpendicular  to  the  base,  bisects  the  vertical  angle,  and  divides 
the  triangle  into  two  symmetrical  right  spherical  triangles. 

By  solving  one  of  these,  we  can  find  the  required  parts  of  the  given 
triangle. 

1.   Given  a  =  115°,  6  =  115°,  C  =  71°  48'  ;  find  A,  B,  and  c. 

Denoting  the  elements  of  one  of  the  right  triangles  by  A',  B',  O,  a', 
V,  and  c',  where  C1  is  the  right  angle,  we  have 

c'  =  a  =  115°,  and  A'  =  }  C  =  35°  54'. 
We  have  then  to  find  the  parts  a'  and  B'  in  this  triangle. 


By  §  140,       sin  A1  =      _,  and  cos  c'  =  cot  A1  cot  B'. 
sin  c' 

-         + 
Or,  sin  a'  =  sin  c'  sin  A\  and  cot  B'  =  cos  c'  tan  A'. 

log  sin  c'  =  9.957276  -  10  log  cos  c'  =  9.625948  -  10 

log  sin  A1  =  9.768173  -  10  log  tan  A'  =  9.859666  -  10 

log  sin  a'  =  9.725449  -  10  log  cot  B1  =  9.485614  -  10 

a'  =  32°  6'  8.6".  180°  -  B'  =  72°  59'  23.5". 

B'  =  107°  0'  36.5". 

Then  in  the  given  isosceles  triangle, 

A  =  B  =  B'  =  107°  0'36.5",  and  c  =  2  a'  =  64°  12'  17.2". 


EXAMPLES. 

Solve  the  following  isosceles  spherical  triangles : 

2.  Given  A=   27°  12',  B=   27°  12',    c  =  135°20'. 

3.  Given    a  =  152°  6',      b  =  152°  6',    C=   67°  46'. 

4.  Given   a  =  112°  25',    b  =  112°  25',    c  =  123°48'. 

5.  Given  A  =  159°,        B  =  159°,         a  =  137°  39'. 


96  SPHERICAL  TRIGONOMETRY. 

XII.    OBLIQUE  SPHERICAL  TRIANGLES. 

GENERAL  PROPERTIES  OF  SPHERICAL  TRIANGLES. 

151.   In  any  spherical  triangle,  the  sines  of  the  sides  are  proportional  to 
the  sines  of  their  opposite  angles. 

C 


Let  ABC  be  any  spherical  triangle,  and  draw  the  arc  CD  perpen- 
dicular to  AB. 

There  will  be  two  cases  according  as  CD  falls  upon  AB  (Fig.  1),  or 
upon  AB  produced  (Fig.  2). 

In  the  right  triangle  ACD,  in  either  figure,  we  have 


Also,  in  Fig.  1,  sin  B  = 

sma 

And  in  Fig.  2,  sin  B  =  sin  (180°  -  CBD) 


sin  a 
Dividing  these  equations,  we  have  in  either  case 

sin  CD 
sin  A       sin  b       sina 


sin  B  ~~  sin  CD  ~~  sin  b 


sn  a 


In  like  manner,  «nB     rinft,  (86) 

sm  C     sin  c 


i  sin  ^±      sin  a  fon\ 

and  — ^  = (87) 

sin  C      sin  c 

152.  In  any  spherical  triangle,  the  cosine  of  any  side  is  equal  to  the 
product  of  the  cosines  of  the  other  two  sides,  plus  the  continued  product  of 
their  sines  and  the  cosine  of  their  included  angle. 


OBLIQUE   SPHERICAL   TRIANGLES.  97 

In  the  right  triangle  BCD,  in  Fig.  1,  §  151,  we  have,  by  (75), 
cos  a  =  cos  BD  cos  CD  =  cos  (c  —  AD)  cos  CD. 

And  in  Fig.  2, 

cos  a  =  cos  BD  cos  CD  =  cos  (AD  —  c)  cos  <7Z>. 

Whence,  in  either  case,  by  (12), 

cos  a  =  cos  c  cos  AD  cos  CD  +  sin  c  sin  AD  cos  CZ>. 

But  in  the  right  triangle  ACD, 

cos  AD  cos  <7Z)  =  cos  6,  by  (75). 

And,         sin  AD  cos  (7Z>  =  sin  AD  cos  6    =  cos  6  tan  J.Z> 


=  sin  b—  -  =  sin  b  cos  A,  by  (77). 
tan  b 

Whence,  cos  a  =  cos  b  cos  c  +  sin  b  sin  c  cos  A.  (88) 

In  like  manner,        cos  b  =  cos  c  cos  a  +  sin  c  sin  a  cos  5,  (89) 

and  cos  c  =  cos  a  cos  b  -f  sin  a  sin  &  cos  (7.  (90) 

153.  Let  ABC  and  A'B'C'  be  a  pair  of  polar  triangles. 


Applying  formula  (88)  to  the  triangle  A'B'C',  we  obtain 

cos  a'  =  cos  b'  cos  c'  -f  sin  6'  sin  c'  cos  -4'. 

Putting  for  a',  6',  c',  and  A'  the  values  given  in  §  131,  6,  we  have 
cos  (180°  -  A)  =  cos  (180°  -  B)  cos  (180°  -  C) 

+  sin  (180°  -  B)  sin  (180°  -  C)  cos  (180°  -  a). 

Whence,    —  cos  A  =  (—  cos  B)  (—  cos  (7)  4-  sin  5  sin  <7(—  cos  a)  (§  33). 

That  is,         cos  A  =  —  cos  B  cos  C  -f  sin  1?  sin  C  cos  a.  (91) 

Similarly,      cos  B  =  —  cos  C  cos  ^4  -|-  sin  C  sin  J.  cos  &,  (92) 

and  cos  (7  =  —  cos  A  cos  J5  +  sin  A  sin  5  cos  c.  (93) 

The  above  proof  illustrates  a  very  important  application  of  the  theory 
of  polar  triangles  to  Spherical  Trigonometry. 


98  SPHERICAL  TRIGONOMETRY. 

If  any  relation  has  been  found  between  the  elements  of  a  spherical 
triangle,  an  analogous  relation  may  be  derived  from  it,  in  which  each  side 
or  angle  is  replaced  by  the  opposite  angle  or  side,  with  suitable  modifica- 
tions in  the  algebraic  signs. 

154.    To  express  the  sines,  cosines,  and  tangents  of  the  half-angles  of  a 
spherical  triangle  in  terms  of  the  sides  of  the  triangle. 
From  (88),  §  152     sin  b  sin  c  cos  A  =  cos  a  —  cos  b  cos  c. 

Whence,  cos  A  =  cos  a  ~  cos  b  cos  c.  (A) 

sin  b  sin  c 

Subtracting  both  members  from  1,  we  have 
1  ,   A  _  1      cos  a  —  cos  b  cos  c     cos  b  cos  c  +  sin  b  sin  c  —  cos  a 

J.  —  COS  -Zl  —  JL ; ; « 

sm  b  sm  c  sin  b  sin  c 

TTT,  cos  (6  —  c)  —  cos  a 

Whence,  by  (31),     2  sm2  \A  = v  .    _  } . 

sin  b  sm  c 

But  by  (20),    cos  y  —  cos  x  =  2  sin  1  (a;  +  y)  sin  £  (#  —  y).  (B) 


Whence,    2Sm^  =  Tc>]  sin^a-(6  ~c>1, 

sin  o  sin  c 

or  sin2  1  A  =  siniO  +  6  -  c)  sin^-(a  -  6  +  c) 

sin  b  sin  c 

Denoting  the  sum  of  the  sides,  a  +  b  -f  c,  by  2  s,  we  have 


and  a-6-f-c=(a  +  6  +  c)-26  =  2s-26 

_,  .   2  sin  (s  —  6)  sin  (s  —  c) 

Whence,  sin2  \  A  —  --  =  —  .  —  /    .    v  --  L  . 

sm  b  sm  c 


.                 /sin  (s  —  6)  sin  (s  —  c)  , 

Or,  sin  4  A  =\\ * — .     '.    .     '-  (94 

2  \  sin  7)  sin  r.  v 


sin  6  sin  c 

TVI  •      i    T>       *  /S^n  (S  —  C)  Sin  (S  ~  a)  /oe\ 

In  like  manner,    sin  1 B  =\/  — *— s — J  .    v ^,  (95) 

'  sm  c  sm  a 

/sin  (s  —  a)  sin  (s  —  b) 

and  smi(7=A/ W-  —£.  (96) 

^  sm  a  sm  o 

Again,  adding  both  members  of  (A)  to  1,  we  have 

cos  a  —  cos  b  cos  c  _  cos  a  —  (cos  b  cos  c  —  sin  b  sin  c) 


1  +  cos  A  =  1 


Whence,  by  (32),    2  cos*  *  4  = 

J  v    " 


sin  6  sin  c  sin  6  sin  c 

cos  a  —  cos  (b  -f-  c) 


sm     sm  c 


OBLIQUE   SPHERICAL  TRIANGLES.  99 


Then  by  (B),    2cos^^  =  C  +  « 

'    J  v   ''  sm  b  sm  c 

Putting  a  +  6  +  c  =  2s,  whence  b  -f-  c  —  a  =  2  (s  —  a),  we  have 

sin  s  sin  (s  —  a) 
=—         r-4  -  '— 
sm  6  sm  c 


_  ^  /sm  s  sm  (s  —  a)  ^  . 

Or,  cos  4  A  =  \ .    .   \ '-  (97) 

*      sm  b  sin  c 


T         Tl  IT.  I8111  S    Sln  (S  —    &)  /00\ 

In  like  manner,     cos  i  .5=  A/—        —^ ^>  (98) 

2  ^       sin  ^  sin  a 


,  .~        /sin  s  sin  (s  —  c)  ,,. 

and  MBiO-  ^.  (99) 


Dividing  (94)  by  (97),  we  have 


.        /sin  (s  —  6)  sin  (s  —  c)     /     sin  6  sin  c 
*  \  sin  ft  sin  c.  \sin  &  sin  (s  — 


sin  6  sin  c  ^  sin  s  sin  (s  —  a) 

(100) 


sm  s  sm  (s  —  a) 


/sin  (s  —  c)  sin  (s  —  a)  /,«,\ 

In  like  manner,     tan  4-  B  =  \  -  5  --  .    /       ^     >  (101) 

^      sm  s  sm  (s  —  b) 


/sin  (s  —  a)  sin  (s  —  6) 


and 


155.    Tb  express  the  sines,  cosines,  and  tangents  of  the  half-sides  of  a 
spherical  triangle  in  terms  of  the  angles  of  the  triangle. 

From  (91),  §  153,  sin  B  sin  (7  cos  a  —  cos  A  +  cosS  cos  C. 


Whence,  COSg  =  C.  (A) 

smBsmO 

Then,  l 


sm  B  sm  C 


0  2  in21     _  —  (cos  ^  cos  C  —  sinjB  sin  C)  —  cos  A 

sin  5  sin  G 


sin  5  sin  (7 

Thenby(19), 

J  v    " 


100  SPHERICAL   TRIGONOMETRY. 

Denoting  the  sum  of  the  angles,  A  +  B  +  (7,  by  2  S,  we  have 


sin  B  sin  C 

Or, 

In  like 

and 
Again, 
1  4-  cos  a  -. 

•    l           1     cos  S  cos  (S  —  A) 

(103) 
(104) 
(105) 

sin  5  sin  (7 

*           sin  B  sin  (7 

mannner      -in  1  fr  —  J     cos  ^  cos  (S  ~  ^ 

^           sin  C  sin  ^4 

•    i    _  .    /     cos  S  cos  (S  —  C) 

^           sin  A  sin  jB 

adding  both  members  of  (A)  to  1,  we  have 
.j      cos  A  +  cos  5  cos  (7     cos  A  +  cos  J5  cos  (7  -h 

Then,  2  cos^a  =  cos  ^1  +  COB  (B  -  O) 

sin  .B  sin  O 


JB  - 


Or, 


sin  jB  sin  C 
~  °  cos     -^  ~ 


sin  S  sin  ( 
-C),  and  .4-  B  +  O 


Whence,  cos^  a  ^  cos  (^ 


sm  B  sm  (7 


Or,  cos  i  a  =\ r*^  ~     '       ^    ~     ;-  (106) 

v  sin  B  sin  (7 


In  like  manner,      eoBJft  =  Jcos(^-  g)cos(f  ~  A),  (107) 

\  sin  r7sin  >4 


and  cosic=-— .  (108, 

^ 

Dividing  (103)  by  (106),  we  have 


cos      - 


Inlikema^er,      tanP  ,  (HO) 


OBLIQUE  SPHERICAL  TRIANGLES.  101 

NAPIER'S  ANALOGIES. 
156.   Dividing  (100)  by  (101),  we  have 

tan  |  A  _    /sin  (s  —  b)  sin  (s  —  c)    I     sing  sin  (s  —  b) 
tan^B  ~~*      sin  s  sin  (s  — a)        *  sin  (s  —  c)  sin  (s  —  a) 

n  sin%Acos%B_    /sin2  (s  —  b)  _  sin  (s  —  b) 

'  cos^A  sin^"  \sin2(s  -  a)  ~  sin(s  -  a)" 

Whence  by  composition  and  division, 

sin -^ .4  cos ^.B  4-  cos \A  sin^-jB     sin(s  —  6)4-  sin(s  —  a) 
sin  \  A  cos  \  B  —  cos  ^  A  sin  ^B     sin  (s  —  6)  —  sin  (s  —  a) 

Then  by  (9),  (11),  and  (21), 

••—  b  4-  s  —  a] 


sin(i  J.  -  !  B)  ~~  tan-J-[»  -b-(s-  a)] 
But  s  —  b  +  s  —  a  =  2s  —  a  —  b  =  c. 


B)          tan  |  c  , 

Whence,  .    ;;  .  --  ^  =  T~   T7"3  —  ix'  (112) 

sm^(A  —  B)     tan^-(a  —  6) 


157.  Multiplying  (100)  by  (101),  we  have 


sin  $  sin  (s  —  a)       ^      smssm(s  —  b) 


„  sin  \  A  sin  \  B  _    /sin2  (s  —  c)      sin  (s  —  c) 

cos^J-Cos^-B"*      sin2s  sins 

Whence  by  composition  and  division, 

—  sin(g  —  c) 


cos  |  A  cos  %  B  4-  sm^-^4sin^^~"  sins  +  sin(s  —  c) 


Or  bv  ten  coBft^  +  jJ*)  _  tanj[s-(s-c)] 

cos  (±A  -  %B)  ~  tan^  [s  4-  s  -  c] 


But  s-j-s  —  c  =  2s  —  c  =  a 


----B)          tan-j-c 
Whence,  ^-jr  -^(-  =  —  -Y^.  (113) 

cos  %(A  —  B)     tan£  (a  4-  b) 

158.  Applying  formula  (112)  to  the  triangle  A'B'C',  in  the  figure  of 
153,  we  obtain 


tan  J(a'  -  b1) 


102  SPHERICAL  TRIGONOMETRY. 


But,     ±(A'  +  JB»)  =  £(180°  -  a  +  180°  -  &)  =  180°  -  J  (a  +  6)  ; 
i(^f  -  .B')  =  1(180°  -  a  -  180°  +  6)=  -  £(a  -  5)  ; 

£c'  =  J(180°  -  C)=  90°  -  £  O; 
and  |(a'  -  &•)=  £(180°  -  A  -  180°  +  5)=  -  £(J.  -  J5). 


Whence      sin[180°-i(a  +  &)]  _    tan(90°-j.Q) 
sin[-  £(a  -  &)]      ~  tan  [-  ±(A  -      ~ 

Therefore,  by  §§  29,  32,  and  33, 


—  sin|-(a  —  6)""  - 

sin^(g  +  6)_ 
sin^  (a  -  6)  "  ta 

In  like  manner,  from  (113),  we  obtain 


Or  _  - 

sin^  (a  -  6)  "  - 


But,  J(a'  +  6*)=  ^(180°  -  -4  +  180°  -  5)  =  180°  -  %(A  +  J5). 


_        tan(90°-iC) 


cos  [-  J(a  -  6)]     ~  tan[180°  -  ^(J.  +  J3)] 
Therefore,  by  §§  29,  32,  and  33, 

co 


^.       cotjO 


159.  The  formulae  exemplified  in  §§  156,  157,  and  158  are  known  as 
Napier's  Analogies.     In  each  case  there  may  be  other  forms  according 
as  other  elements  are  used. 

SOLUTION  OP  OBLIQUE  SPHERICAL  TRIANGLES. 

160.  In  the  solution  of  oblique  spherical  triangles,  we  may  distinguish 
six  cases  :  \ 

1.  Given  a  side  and  the  adjacent  angles. 

2.  Given  two  sides  and  their  included  angle. 

3.  Given  the  three  sides. 

4.  Given  the  three  angles. 

5.  Given  two  sides  and  the  angle  opposite  to  one  of  them. 

6.  Given  two  angles  and  the  side  opposite  to  one  of  them. 


OBLIQUE   SPHERICAL   TRIANGLES.  103 

By  application  of  the  principles  of  §  131,  6,  the  solution  of  an  exam- 
ple under  Case  2,  4,  or  6,  may  be  made  to  depend  upon  the  solution  of  an 
example  under  Case  1,  3,  or  5,  respectively  ;  and  vice  versa. 

Hence,  it  is  not  essential  to  consider  more  than  three  cases  in  the 
solution  of  oblique  spherical  triangles. 

The  student  must  carefully  bear  in  mind  the  remarks  made  in  §§  146 
and  147. 

161.   CASE  I.     Given  a  side  and  the  adjacent  angles. 

1.   Given  A  =  70°,  B  =  132°,  c  =  116°;  find  a,  6,  and  O. 

By  Napier's  Analogies  (§§  156,  157),  we  have 


sin  %(B  +  A)_       tan^c  cos  $(B  +  A)_       tan^c 

sin  %(B  —  A)~~  tan  i  (6  -  a)'  3        cos  £  (B  —  A)  ~  tan  £  (6  +  a)' 


Whence,  tan  J  (b  —  a)  =  sin  %(B  —  A)  esc  J  (B  +  A)  tan  |  c, 

-f  + 

and  tan  J  (6  +  a)  =  cos£(J3-.4)  sec  £(£  -f  J.)  tan  \  c. 

From  the  data,  ±(B-A)  =  31°,  £  (5  +  A)  =  101°,  £  c  =  58°. 

log  sin  $(B-A)  =  9.711839  -  10        log  cos  %(B  -  A)  =  9.933066  -  10 
log  esc  i  (B  +  ^)  =  0.008053  log  sec  {•  (B  +  A)  =  0.719401 

log  tan  i  c  =  0.204211  log  tan  f  c  =  0.204211 

log  tan  \  (b  -  a)  =  9.924103  -  10          log  tan  |  (6  +  a)  =  0.856678 

|(6  -  a)  =  40°  1'  7.7".  180°  -  £  (6  +  a)  =  82°  4'  51.8". 

|(6  -fa)  =  97°  55'  8.2". 

Then,  a  =  £  (6  +  a)  -  }  (6  -  a)  =  57°  54'  0.5", 

and  6  =  J  (6  +  a)  +  £  (6  -  a)  =  137°  56'  15.9". 

To  find  C,  we  have  by  §  158, 


sin 


tan  t  CB-4>=sm  |  (6+0)  esc  J  (6-a)  tan  J  (B- 
— 


log  sin  I-  (6  +  a)  =  9.995839  —  10 
log  esc  J  (6  -  a)  =  0.191763 
log  tan  J  (J3  -  4)  =  9.778774  -  10 

log  cot  i  O  =  9.966376  -  10 

|  O=  47°  12'  56.7",  and  (7  =  94°  25'  53.4  '. 

Note  1.     The  value  of  G  may  also  be  found  by  the  formula 


Note  2.     The  triangle  is  possible  for  any  values  of  the  given  elements. 


104  SPHERICAL  TRIGONOMETRY. 

EXAMPLES. 

Solve  the  following  spherical  triangles  : 

2.  Given  A  =  78°,          B  =  41°,          c  =  108°. 

3.  Given  5=115°,         (7=50°,          a  =  70°20f. 

4.  Given  A  =  31°  40',     C  =  122°  20',  b  =  40°  40'. 

5.  Given  A  =  108°  12',  B  =  145°  46',  c  =  126°  32'. 

162.  CASE  II.  Given  two  sides  and  their  included  angle. 
1.  Given  b  =  138°,  c  =  116°,  A  =  70°  ;  find  B,  C,  and  a. 
By  Napier's  Analogies  (§  158),  we  have 


sin  %  (b  -f  c)  _        cot  £  ^4.  ,   cos  £  (b  4-  c)  _        cot  %  A 

~(7)>  a        cos  \  (b  -  c)  ~~  tan  £  (.B  +  (7) 


Whence,  tan  £  (5  -  <7)  =  sin  |  (&  -  c)  esc  |(6  +  c)  cot  £  .4, 


and  tan  %  (B  +  O)  =  cos  £  (6  —  c)  sec  £  (6  +  c)  co 

From  the  data,  |(6  -  c)  =  11°,  |(6  +  c)  =  127°,  |  ^1  =  35°. 


log  sin  J  (6  -  c)  =  9.280599  -  10  log  cos  |  (6  -  c)  =  9.991947  -  10 

log  esc  J  (6  +  c)  =  0.097651  log  sec  1  (6  +  c)  =  0.220537 

log  cot  |  ^  =  0.154773  log  cot  £  .4  =  0.154773 


log  tan  £  (J5  -  (7)  =  9.533023  -  10        log  tan  ±(B  +  C)  =  0.367257 

£  (J5  -  O)  =  18°  50'  24.7".          180°  -±(B+C)  =  66°  46'  1.2". 


Then,  B  =  |  (J5  +  C)  +  £  (-B  -  C)  =  132°  4'  23.5", 

and  C  -  £  (B  +  C)  -  1  (B  -  C)  =  94°  23'  34.1". 

To  find  a,  we  have  by  §  156, 


tan  i-  a=  tan  £  (6-c)=sin  £  (B+C)  esc  $  (B- 

sin  -j  ^.o  —  \j  ) 

log  sin  |(5  +  C)  =  9.963272  -  10 
log  esc  |(5  -  O)  =  0.490892 
log  tan  |(6  -  c)  =  9.288652  -  10 

log  tan  \  a  =  9.742816  -  10 

la  =  28°  56'  51.6",  and  a  =  57°  53  '43.2". 
Note.    The  triangle  is  possible  for  any  values  of  the  given  elements. 


OBLIQUE  SPHERICAL  TRIANGLES.  105 

EXAMPLES. 

Solve  the  following  spherical  triangles  : 

2.  Given  a  =  72°,          b  =  47°,  C  =  33°. 

3.  Given  a  =  98°,          c  =  60°,  B  =  110°. 

4.  Given  b  =  70°  40',     c  =  120°20',  A  =  50°. 

5.  Given  a  =  125°  10',  6  =  153°  50',    C  =  140°  20'. 

163.   CASE  III.     Given  the  three  sides. 

The  angles  may  be  calculated  by  the  formulae  of  §  154. 

If  all  the  angles  are  to  be  computed,  the  tangent  formulae  are  the  most 
convenient,  since  only  four  different  logarithms  are  required.  If  but  one 
angle  is  required,  the  cosine  formula  will  be  found  to  involve  the  least  work. 

The  triangle  is  possible  for  any  values  of  the  data,  provided  that  no 
side  is  greater  than  the  sum  of  the  other  two,  and  that  the  sum  of  the 
sides  is  less  than  360°  (§  131,  1  and  3). 

If  all  the  angles  are  required,  and  the  tangent  formulae  are  used,  it 
is  convenient  to  modify  them  as  follows. 


By  (100),       tan  *  A  =  J«h  (•  -  «)  sin  (.  -  6)  sin  (.  -  c) 

™  sm  s  sm2  (s  —  a) 


_         1  /sin  (s  —  a)  sin  (s  —  b)  sin  (s  —  c) 

sin(s-a)  \~  sins 


Denoting  (»  -  a)  sin  («  -  6)  sin  fr  -  c)  b 

sin  s 


sin  (s  —  a) 

k  k 

In  like  manner,    tan  \  B  = ,  and  tan  i  C  = • 

sin  (s  —  b)  sm  (s  —  c) 

1.   Given  a  =  57°,  b  =  137°,  c  =  116° ;  find  A,  B,  and  C. 

Here,                              2s  =  a  +  b  +  c  =  310°. 

Whence,    s  =  155°,  s  -  a  =  98°,  s-b  =  18°,  s  -  c  =  39°. 

log  sin  (s  -  a)  =  9.995753  -  10  log  k  =  9.829330  - 10 

log  sin  (s  -  b)  =  9.489982  - 10  log  sin  (s  -  b)  =  9.489982  -  10 

log  sin  («-«)=  9.798872  - 10  log  tan  1 B  =  0.339348 

log  esc  s  =  0.374052  £  B  =  65°  24'  10.4". 

2)19.658659  -~20  B  =  130°  48'  20.8." 

log  k  =  9.829330  - 10  log  k  =  9.829330  -  10 

log  sin  (s  -  a)  =  9.995753  - 10  log  sin  (s  -  c)  =  9.798872  - 10 

log  tan  i  A  =  9.833577  -  10  log  tan  £  C  =  0.030458 

i  A  =  34°  16'  52.5".  \  C  =  47°  0'  27.0". 

A  =  68°  33'  45.0".  C=  94°  0'  54.0". 


106  SPHERICAL  TRIGONOMETRY. 

EXAMPLES. 

Solve  the  following  spherical  triangles : 

2.  Given  a  =  38°,         5  =  42°,        c  =  61°. 

3.  Given  a  =  101°,       6  =  49°,        c  =  60°. 

4.  Given  a  =  126°,       6  =  152°,      c  =  75°. 

5.  Given  a  =  62°  20',  b  =  54°  10',  c  =  97°  50' ;  find  A. 

164.   CASE  IV.     Given  the  three  angles. 

The  sides  may  be  calculated  by  the  formulae  of  §  155. 

If  all  the  sides  are  to  be  computed,  the  tangent  formulae  are  the  most 
convenient,  since  only  four  different  logarithms  are  required.  If  but  one 
side  is  required,  the  sine  formula  will  be  found  to  involve  the  least  work. 

The  triangle  is  possible  for  any  values  of  the  data,  provided  that  the 
sum  of  the  angles  is  between  180°  and  540°  (§  131,  4),  and  that  each  of 
the  quantities  B+C  —  A,  C  +  A—  B,  and  A  +  B  —  G  is  between  180° 
and  - 180°  (§  134). 

For  such  values  of  the  angles,  S  is  between  90°  and  270°,  and  each  of 
the  quantities  S  —  A,  S  -  B,  and  8—C  between  90°  and  -90°;  then, 
cos  8  is  — ,  while  the  cosines  of  S  —  A,  S  —  B,  and  S  —  C  are  +  (§  20). 

Hence,  the  expressions  under  the  radical  signs  in  the  formulae  are 
essentially  positive,  and  no  attention  need  be  paid  to  the  algebraic  signs. 

If  all  the  sides  are  required,  and  the  tangent  formulae  are  used,  it  is 
convenient  to  modify  them  as  follows : 


cos  S  cos2  (S  -  A) 
By  (109), 


=  cos  (S  -  4)  J_ 


cos  S 


cos  (S  -  A)  cos  (8-B)  cos  (S  -  C) 


Denoting  J ^^. by  K, 

*     cos  (S  -  A)  cos  (S  -  B)  cos  (S  -  C) 

we  have  tan  %a=  Kcos(S  —  A). 

In  like  manner,  tan  £  b  =  TTeos  (S  —  B),  and  tan  £  c  =  Kcos  (S  -  (7). 

1.    Given  A  =  150°,  B  =  131°,  C  =  115° ;  find  a,  6,  and  c. 

Here,  2S  =  A  +  B+C=  396°. 

Whence,    S  =  198°,  S  -  A  =  48°,  S  -  B  =  67°,  S  -  C  =  83°. 


OBLIQUE   SPHERICAL  TRIANGLES.  107 

log  cos  S  =  9.978206  - 10  log  K=  0.737462 

log  sec  (S-A)  =  0.174489  log  cos  (S  -  B)  =  9.591878  - 10 
log  sec  (S-B)  =  0.408122  log  tan  1 5  =  0.329340 

log  sec  (S  -  C)  =  0.914106  |  d  =  64°  53'  58.0". 

2)1.474923  b  =  129°  47'  56.0". 

log  K=  0.737462  log  K=  0.737462 

log  cos  (S-A)  =  9.825511  -  10  log  cos  (S  -  C)  =  9.085894  -  10 

log  tan  |  a  =  0.562973  log  tan  |  c  =  9.823356  -  10 

i-  a  =  74°  42'  4.8".  |  c  =  33°  39'  23.1". 

a  =  149°  24'  9.6".  c  =  67°  18'  46.2". 

Note  1.     By  §  35,  cos  198°  =  -  sin  108°  =  -  cos  18°  ;  whence,  without  regard  to 
algebraic  sign,  log  cos  198°  =  log  cos  18°. 

2.  Given  ^1  =  123°,  5  =  45°,  O=58°;  find  a. 

By  (103),          —  2 v 

*  sin  B  sin  C 

Here,  2S  =  A  +  B+C=  226° ;  whence,  £=113°,  and  S  -  A  =  - 10°. 

log  cos  S  =  9.591878  -  10 

log  cos  (S-A)  =  9.993351  -  10 
log  esc  B  =0.150515 

log  esc  C  =  0.071580 

2)19.807324^20 
log  sin  £  a  =  9.903662  - 10 

%a  =  53°  13' 51.3",  and  a  =  106° 27' 42.6". 

Note  2.    By  §  29,  cos  (-  10°)  =  cos  10°. 

EXAMPLES. 
Solve  the  following  spherical  triangles : 

3.  Given  ^1  =  74°,  5  =  82°,  C=67°. 

4.  Given  .4  =  120°,  5=130°,  O=140°. 

5.  Given  ^[  =  138°  16',      B  =  33°  11',  C=36°53'. 

6.  Given  A  =  91°  10',        B  =  85°  40',  C  =  78°  30' ;  find  5. 

165.  CASE  V.     Given  two  sides  and  the  angle  opposite  to  one  of  them. 
1.   Given  a  =  58°,  6  =  137°,  B  =  131° ;  find  A,  0,  and  c. 


By  (85),  =         ,  or  sin  A  =  sin  a  esc  6  sin  B. 

J  ^    '          sin  B     sin  b 


108  SPHERICAL   TRIGONOMETRY. 

log  sin  a  =  9.928420  -  10 
log  esc  b  =  0.166217 
log  sin  B  =  9.877780  -10 

log  sin  .4  =  9.972417 -10 

A  =  69°  47'  41.6",  or  110°  12'  18.4"  (§  147). 
To  find  C  and  c,  we  have  by  §§  156  and  158, 

cot  J  C=  sin-*-  (b  +  a)  csci  (b  -  a)  tan£  (B  -  A), 
and  tan^c  =  sin£ (B  +  A)  csc%(B  -  A)  tan|(6  -  a). 

Using  the  first  value  of  A,  we  have 

±(B  +  A)  =  100°  23' 50.8",  and  J (B-A)  =  30° 36' 9.2". 
Also,  |  (6  +  a)  =  97°  30',  and  J  (6  -  a)  =  39°  30'. 

log  sin  J  (6  +  a)  =  9.996269  -  10  log  sin  ±(B  +  A)  =  9.992810  -  10 

log  esc  |  (b  -  a)  =  0.196489  log  esc  |  (B  -  -4)  =  0.293214 

log  tan  %(B  -  A)  =  9.771924  -  10  log  tan  1  (&  -  a)  =  9.916104  -  10 

log  cot  J  (7  =  9.964682  -*• 10  log  tan  £  c  =  0.202128 

|  (7=47°  19' 37.8".  |c=   57°  52' 35.0". 

O=  94°  39f  15.6".  c  =  115°  45'  10.0". 

Using  the  second  value  of  A,  we  have 

^ (5  +  ^)  =120° 36' 9.2",  and  ±(B-A)=  10° 23' 50.8". 
log  sin  $  (b  +  a)  =  9.996269  -  10  log  sin  $(B  +  A)=  9.934861  -  10 

log  esc  i  (6  -  a)  =  0.196489  log  esc  1  (B  -  A)  =  0.743583 

log  tan  \  (B  -  A)  =  9.263608  -  10  log  tan  1  (b  -  a)  =  9.916104  - 10 

log  cot  $  C  =  9.456366  -  10  log  tan  £  c  =  0.594548 

i(7=    74°  2' 22.1".  £c=    75°  43' 43.6". 

O=  148°  4'  44.2".  c  =  151°  27'  27.2". 

Thus,  the  two  solutions  are : 

1.  A=    69°  47' 41.6",  C=    94°  39' 15.6 ",  c  =  115°  45 '10.0". 

2.  ^4=110°  12' 18.4",  (7=148°   4' 44.2",  c  =  151°  27' 27.2". 

As  in  the  corresponding  case  in'  the  solution  of  oblique  plane  triangles 
(compare  §§  117  to  120),  there  may  sometimes  be  two  solutions,  sometimes 
only  one,  and  sometimes  none,  in  an  example  under  Case  V. 

After  the  two  values  of  A  have  been  obtained,  the  number  of  solutions 
may  be  readily  determined  by  inspection ;  for,  by  §  131,  2,  if  a  is  <  6, 
A  must  be  <  B ;  and  if  a  is  >  6,  A  must  be  >  B. 

Hence,  only  those  values  of  A  can  be  retained  which  are  greater  or  less 
than  B  according  as  a  is  greater  or  less  than  b. 


OBLIQUE  SPHERICAL   TRIANGLES.  109 

Thus,  in  Ex.  1,  a  is  given  <  b  ;  and  since  both  values  of  A  are  <  B,  we 
have  two  solutions. 

Again,  if  the  data  are  such  as  to  make  log  sin  A  positive,  there  will  be 
no  solution  corresponding. 

2.   Given  a  =  58°,  c  =  116°,  C  =  94°  50';  find  A. 


In  this  case,     ^       =  !EL^   Or  sin  A  =  sin  a  esc  c  sin  C. 
sin  C     sin  c 

log  sin  a  =  9.928420  -  10 
log  esc  c==  0.046340 
log  sin  0=9.998453  -10 

log  sin  .4=9.973213  -10 

^  =  70°  4'  57.1",  or  109°  55'  2.9". 

Since  a  is  given  <  c,  only  values  of  A  which  are  <  C  can  be  retained  ; 
then  the  only  solution  is  A  =  70°  4'  57.1". 

3.   Given  b  =  126°,  c  =  70°,  B  =  56°;  find  C. 

In  this  case,    ^1^  =  ^Bj?   Or  sin  <7  =  sin  c  esc  &  sin  B. 
sin  .B     sin  6 

log  sin  c  =  9.972986  -  10 

log  esc  b  =  0.092042 

log  sin  B  =  9.918574  -  10 

log  sin  C  =  9.983602  -  10 

(7  =  74°  21'  13.8",  or  105°  38'  46.2". 

Since  both  values  of  C  are  >  B,  while  c  is  given  <  &,  there  is  no  solu- 
tion. 

EXAMPLES. 
3olve  the  following  spherical  triangles  : 

4.  Given  6=    99°  40',     c=   64°  20',  B=   95°  40'. 

5.  Given  a  =   40°,          6  =  118°  20',  A=   29°  40'. 

6.  Given  a  =  115°  20',     c=146°20',  O=141°10'. 

7.  Given  a  =  109°  20',  c=  82°  1'8",  ^  =  107°  40'. 

8.  Given  b  =  108°  30',  c=  40°  50',  (7=  39°  50'. 

9.  Given  a  =  162°  20',  6=  15°  40',  5  =  125°. 
10.  Given  a  =  55°,    c  =  138°10',  A=  42°  30'. 


110  SPHERICAL  TRIGONOMETRY. 

166.   CASE  VI.  Given  two  angles  and  the  side  opposite  to  one  of  them. 
1.   Given  A  =  110°,  B  =  122°,  b  =  129°  ;  find  a,  c,  and  C. 


By  (85),  =      -      or  sin  a  =  sin  A  esc  B  sin  6. 

sm  b      sin  B 

log  sin  .4  =  9.972986  -10 
log  csc  J3  =  0.071580 
log  sin  6  =  9.890503  -10 

log  sin  a  =  9.935069  -  10 

a  =  59°  26'  37.6",  or  120°  33  '  22.4"  (§  147). 
To  find  c  and  <7,  we  have  by  §§  156  and  158, 

tan  i  c  =  sin  %(B  +  A)  esc  %(B-A)  tan  £  (b  -  a), 
and  cot  %  C  =  sin  ^  (b  +  a)  esc  £  (b  —  a)  tan  ^-  (B  —  A). 

Using  the  first  value  of  a,  we  have 

£(6  +  a)=94°13'18.8",  and  J(6  -  a)=  34°46'41.2". 
Also,  ±(B  +  A)  =  116°,  and  ±(B-A)=  6°. 

log  sin  ±(£  +  A)=  9.953660  -  10  log  sin  £  (b  +  a)  =  9.998820  -  10 

log  esc  %(B-A)=  0.980765  log  esc  J  (6  -  a)  =  0.243821 

log  tan^  (6  -  a)=  9.841642-10  log  tan£(£-  ^)=  9.021620  -  10 

log  tan  £  c  =  0.776067  log  cot  J  C  =  9.264261  -  10 

Jc=   80°  29'  34.8".  |(7  =    79°  35'  14.1". 

c  =  160°59'   9.6".  (7=159°  10'  28.2". 

Using  the  second  value  of  a,  we  have 

|(6  +  a)  =  124°  46'  41.2",  and  £  (b  -  a)  =  4°  13'  18.8". 
log  sin  %(B  +  A)=  9.953660  -  10  log  sin  £  (6  +  a)  =  9.914537  -  10 

log  esc  \  (B  -A)=  0.980765  log  esc  \  (b  -  a)  =  1.133009 

log  tan  \  (b  -  a)  =  8.868171  -  10  log  tan  ±(B-A)=  9.021620  -  10 

log  tan-Jc  =  9.802596  -  10  log  cot  £  C  =  0.069166 

J  c  =  32°  24'  17.8".  \  C  =  40°  27'  24.1". 

c  =  64°  48'  35.6".  C=  80°  54'  48.2". 

Thus  the  two  solutions  are  : 

1.  a=   59°  26'  37.6",  c  =  160°59'   9.6",  (7=  159°  10  '28.2". 

2.  a  =  120°  33'  22.4",  c=    64°  48'  35.6",  C=   80°  54'  48.2". 

In  examples  in  Case  VI.,  as  well  as  in  Case  V.,  there  may  sometimes 
be  two  solutions,  sometimes  only  one,  and  sometimes  none. 


OBLIQUE   SPHERICAL  TRIANGLES.  HI 

As  in  Case  V.,  only  those  values  of  a  can  be  retained  which  are  greater  or 
less  than  b  according  as  A  is  greater  or  less  than  B. 

Also,  if  log  sin  a  is  positive,  the  triangle  is  impossible. 

EXAMPLES. 

Solve  the  following  spherical  triangles : 

2.  Given  B  =  116°,          C  =   80°,  c=   84°. 

3.  Given  ^  =  132°,          5  =  140°,  6  =  127°. 

4.  Given  A=   62°,          C  =  101°  58'  24",  a=    64°  30'. 

5.  Given  A  =  133°  50',    B=   66°  30',  a=   81°  10'. 

6.  Given  B=   22°  20',     C  =146°  40',  c  =  138°20'. 

7.  Given  A=   61°  40',     0=140°  20',  c  =  150°20'. 

8.  Given  5=   73°,           0=    81°  20',  6  =  122°  40'. 

APPLICATIONS. 

167.  In  problems  concerning  navigation,  the  earth  may  be  regarded  as 
a  sphere. 

The  shortest  distance  between  any  two  points  on  the  surface  is  the  arc 
of  a  great  circle  which  joins  them ;  and  the  angles  between  this  arc  and 
the  meridians  of  the  points  determine  the  bearings  of  the  points  from  each 
other. 


E 


Thus,  if  Q  and  Q'  are  the  points,  and  PQ  and  PQ'  their  meridians,  the 
angle  PQQ'  determines  the  bearing  of  Q'  from  Q,  and  the  angle  PQ'Q 
determines  the  bearing  of  Q  from  Q'. 

If  the  latitudes  and  longitudes  of  Q  and  Q1  are  known,  the  arc  QQ' 
and  the  angles  PQQ'  and  PQ'Q  may  be  determined  by  the  solution  of  a 
spherical  triangle. 

For  if  EE'  is  the  equator,  and  PG  the  meridian  of  Greenwich,  we  have 

^  QPQ'  =Z  Q'PG  -Z.  QPG  =  longitude  Q'  -  longitude  Q. 


112 


SPHERICAL  TRIGONOMETRY. 


Also, 


PQ  =  PE  -  QE   =  90°  -  latitude  Q, 
and  PQ'  =  PE'  +  Q'E'  =  90°  +  latitude  Q'. 

Thus,  in  the  spherical  triangle  PQQ',  two  sides  and  their  included 
angle  are  known,  and  the  remaining  elements  may  be  computed. 

When  QQ'  has  been  found  in  degrees,  its  length  in  miles  may  be 
calculated  by  rinding  the  ratio  of  its  arc  to  360°,  and  multiplying  the 
result  by  the  length  of  the  circumference  of  a  great  circle ;  in  the  following 
problems,  the  radius  of  the  earth  is  taken  as  3956  miles. 

EXAMPLES, 

1.  Boston  lies  in  lat.  42°  21'  K,  Ion.  71°  4'  W. ;  and  the  latitude  of 
Greenwich  is  51°  29'  N.     Find  the  shortest  distance  in  miles  between  the 
places,  and  the  bearing  of  each  place  from  the  other. 

2.  Calcutta  lies  in  lat.  22°33'K,  Ion.  88°  19'  E. ;  and  Valparaiso  in 
lat.  33°  2'  S.,  Ion.  71°  42'  W.     Find  the  shortest  distance  in  miles  between 
the  places,  and  the  bearing  of  each  place  from  the  other. 

3.  Sandy  Hook  lies  in  lat.  40°  28'  K,  Ion.  74°  1'  W. ;  and  Queenstown 
in  lat.  51°  50'  K,  Ion.  8°  19'  W.     In  what  latitude  does  a  great  circle 
course  from  Sandy  Hook  to  Queenstown  cross  the  meridian  of  50°  W.  ? 

168.  The  Astronomical  Triangle, 


Let  0  be  the  position  of  an  observer  on  the  surface  of  the  earth ;  P  the 
celestial  north-pole ;  Z  the  zenith. 

The  great  circle  EE',  having  P  for  its  pole,  is  called  the  celestial  equa- 
tor; and  the  great  circle  HH',  having  Z  for  its  pole,  is  called  the  horizon. 

Let  S  be  the  position  of  a  star ;  PSM  a  meridian  passing  through  5; 
ZSN  a  quadrant  of  a  great  circle  passing  through  Z  and  S. 

The  arc  SM  is  called  the  declination  of  the  star ;  and  is  called  declina- 
tion north  or  south,  according  as  the  star  is  north  or  south  of  the  celestial 
equator. 

The  angle  SPZ  is  called  the  hour-angle  of  the  star ;  the  arc  SN  is 
called  its  altitude;  the  angle  PZS,  its  bearing  or  azimuth. 


ANSWEES. 


12.  85°  56' 37.32". 

13.  14°  19' 26.22". 


3.     UK,    2W7T±^- 
O 

4.   (2  n +  !)=,« 


5. 


6. 


§  56;  page  30. 

14.  95°  29' 34.8". 

15.  20°  27' 2.52". 

§  63 ;  page  39. 
7. 

.)W1T  8. 

9. 

10. 


16.   130°  55' 5.952" 


6 

±tan-1(iV7l 


§  76; 

page  44. 

2. 

1 

.5441. 

6. 

2.1003. 

10. 

2.5104. 

14. 

3.4192. 

3. 

1 

.6990. 

7. 

2.2922. 

11. 

2.5774. 

15. 

3.7814. 

4. 

1 

.6232. 

8. 

2.3892. 

12. 

2.9421. 

16. 

4.0794. 

5. 

1 

.8751. 

9. 

2.3222. 

13. 

2.8363. 

17. 

4.2006. 

§  78 ;  page  44. 

2.  0.5229.  5.   1.1549.  8.   0.2831.  11.  1.4592. 

3.  0.2431.  6.  0.2589.  9.   0.7939.  12.  1.3468. 

4.  1.6532.  7.  2.3522.  10.  2.1303.  13.  2.0424 

§81;  page  45. 


3.  3.3397. 

8.  0.5663. 

13.  0.6171. 

19.  0.8752. 

4.  1.7475. 

9.  0.0430. 

14.  0.29i8. 

20.  0.0794. 

5.  0.6338. 

10.  0.1165. 

16.  0.0495. 

21.   0.4248. 

6.  8.6826. 

11.  0.0939. 

17.  0.0365. 

22.  0.1051. 

7.   1.0460. 

12.  0.5440. 

18.   0.7007. 

23.  0.0406. 

119 


120 


ANSWERS. 


§  85 ;  page  47. 

2.  0.5562.              5.   8.9912-10.      8.   8.5932-10.  11.  2.3064 

3.  1.0491.             6.   7.5353-10.      9.   6.6074-10.  12.  0.1151. 

4.  9.9242-10.     7.  3.4592.             10.   9.2885-10.  13.  0.7782. 


4.  0.011739. 

5.  2.527511. 

6.  6.780210-10. 

7.  4.812917.- 

8.  3.960116. 

9.  7.013152-10. 


§  86;  page  47. 

10.  4.942550-10. 

11.  5.863566. 

12.  5.640409-10. 

15.  6.61005. 

16.  55606.5. 

17.  .0110890. 

24.  .00000130514. 


18.  186.334. 

19.  .00223905. 

20.  .0000100006r 

21.  9776.67. 

22.  467929. 

23.  .000342770, 


§  91  ;  pages  50,  51. 

1.  1897.85. 

17. 

244.004. 

35.  .695490. 

2.  -193315. 

18. 

.00279116. 

36.  .542699. 

3.  .309170.. 

19. 

.000000237177. 

37.  -36.0189. 

4.  .00110375. 

20. 

2.23607. 

38.  -11.1122. 

5.  6.36103. 

21. 

1.14870. 

39.  .943241. 

6.  .0301742. 

22. 

-  1.22028. 

40.  2.62762. 

7.  31.2004. 

23. 

1.77828. 

41.  2.53217. 

8.  -.132693. 

24. 

.668289. 

42.  -1.79616. 

9.  .126965. 

25. 

.645831. 

43.  1.03242. 

10.  .0235770. 

26. 

.137751. 

44.  .298557. 

11.  -1.16493. 

27. 

-  .370134. 

45.  .0448607. 

12.  -.00256105. 

30. 

13.8289. 

46.  .794509. 

13.  3692.77. 

31. 

2.48722. 

47.  1.80492. 

14.  .277996. 

32. 

1.05557. 

48.  179.596. 

15.  -15896.0. 

33. 

.0000214279. 

49.  1.88270. 

16.  .0316228. 

34. 

.00710469. 

50.  .000193152, 

51.  - 

.0995935. 

52. 

1.34384. 

§ 

92:  page  52. 

3.  a  -.2831+. 

8  x 

3  log  a 

4.  »  =  -2.173+. 

O.  Jj  — 

4  log  n  —  2  log  m 

5.  x  =1.155+. 

Q       /•£  

1 

6.  a;  =  -.1765+. 

2 

10.  x  = 

1  or  —5. 

log  a  —  2  log  b 


ANSWERS. 


121 


§  93  ;  page  52. 
2.  3.7004+.    3.  -.06546+.    4.  -6.059+. 

6.  -.4601+.   7.  .3494+.   9.  4. 


10-r 


1.  9.345950-10. 

2.  0.376890. 

3.  9.932630-10. 

4.  9.865995-10. 

5.  9.243533-10. 

6.  9.163433.  - 10. 


1.  .68573. 
2  .25232. 
3.  .06344. 


1.  8.338076-10. 

2.  8.810945-10. 


§  94  ;  page  53. 

7.  0.302190. 

8.  0.153906. 

9.  0.002256. 

10.  59°  15' 26.4". 

11.  33°  0' 16.1". 

12.  81°  7' 37.9". 

§  95  ;  page  53. 

4.  .69518. 

5.  .92163. 

6.  .86962. 

10.   29°  9' 13.8". 

§  96 ;  page  53 

3.  1.369926. 

4.  0°  58' 51.06". 


5.  3.326+. 

-i  12  | 

6       o 


13.  27°  31 '50.5". 

14.  8°  41' 32.7". 

15.  75°  45' 9.8". 

16.  49°  38' 57.1". 

17.  23°  26' 30.9". 


7.  51°  36' 42.9". 

8.  15°  28' 22.5". 
9.- 66°  14' 40.0". 


5.  0°  24' 53.79". 

6.  1°  37 '41.93". 


102 ;  pages  56  to  58. 


l.o=    1.8117,  6  =    6.7615.  14.  a 

2.  6  =  11.7793,  c  =  12.7965.  15.  a 

3.  a  =  16.7820,  c  =  26.1081.  16.  a 
t.A  =  34°  22'  7.1",  b  =  .511764.  17.  A 

5.  A  =  33°  8'  56.3",  c  =  499.252.  18.   b 

6.  b  =10.3547,  c  =  13.1404.  19.  A 

7.  a  =  .0036235,  6  =  .013523.  20.    b 
S.A  =  39°  49'  24.6",  a  =  48.8645.  21.   a 
9.  a  =  148.407,  c  =  948.680.  22.   b 

10.  A  =  49°  53'  54.9",  c  =  4.46330.  23.  A 

11.  b  =  .000336374,   c=. 00336715.  24.   a 

12.  a  =  3821.55,  6  =  3641.34.  25.  a 

13.  A  =  35°  53'  55.2",  6  =  731.237.  26.  a 


=  176.533,  c  =  191.993. 
=  20455.6,  c  =  21405.6. 
=  2.40989,  6  =  .812578. 
=  19°  31' 57.2",  c=  .000505172 
=  77.6330,  c  =  91.2952. 
=  32°  10 '16.5",  a  =  388.471. 
=  644.109,  c  =  650.272. 
=  34308.0,  6  =  23381.6. 
=  4.48174,  c  =  8.5085. 
=  39°  21' 54.1",  6  =  121.240. 
=  .00247181,  c  =  .00360016. 
=  16001.6,  c  =  85725.1. 
=  3624500,  6  =  8821960. 


122 


ANSWERS. 


27.  A  =  76°  33'  49.0",  a  =  24234.4. 

28.  a  =  207302,  b  =  421170. 


29.  a  =  .507624,  c  =  .525355. 

30.  A  =  60°  14'  12.9",  c  =  774.563. 


36.  a  =  4925.31. 

37.  20.573. 

38.  83.271ft. 

39.  31°  47 '24.5". 

40.  36°  37' 58.0". 


41.  99.4565  mi. 

42.  10.2352. 

43.  19°  49' 46.7". 

44.  365.64ft. 

45.  56°  18' 35.7". 


31.  c  =  252.103. 

32.  a  =  1.73561. 

33.  c  =  122748. 

34.  J.  =  47°42'47.8". 

35.  a  =  .344647. 

46.  25.2230  mi.,  30.0750  mi.       48.  14.4853,  15.6787. 

47.  21.6514.  49.  517.51ft. 

50.  17.2624.         51.  420.867ft.       52.  437.605. 

53.  10.392.  54.  482.1ft. 

55.   Kate,  6.79668  miles  an  hour;  bearing,  1ST.  63°  8' 28.5"  W. 


2.  J3  =  89°59'42.8". 

3.  jB  =  89°23'22.6". 

4.  ^1  =  89°  59' 37.2". 


2.  6.9066. 

3.  .151079. 

4.  5699.7. 


§  104 ;  page  60. 

5.  B  =  89°  59'  59.0". 

6.  ^1  =  89°  43' 13.6". 


§  106 ;  page  61. 

5.  .089433. 

6.  8130.9. 

7.  .0067825. 


8.  2.18876. 

9.  107.762. 
10.   .0487840. 


2.  6  =  283.331,    c 

3.  a  =  .340132,    c  =  . 986084. 

4.  a  =  29.0595,    6  =  18.3742. 

5.  a  =  .0313440,  c  =  .0498733. 

6.  6  =  5.76721,     c  =  2.16917. 


§  114;  page  67. 

267.677.                7.  a  =  5058.5,    6  =  3683.53. 

8.  a  =  .299674,  6  =  .731538. 

9.  a  =  4.01036,  c  =  3.55195. 
10.  6  =  56719.9,  c  =  23073.5. 


115 ;  pages  68,  69. 


2.  .4=118°  17' 57.4",  6=44.7274. 

3.  A=  60°  44' 39.5",  c=965.282. 

4.  <7  =  63° 49'   9.3",  a=4.48237. 

5.  B=  28°  43' 49.0",  c  =  1.44246. 

6.  5=145°  35' 24.7",  a  =  1045.74. 


7.  0=63°  48' 28.1",    6=13.7387, 

8.  ^=67°  55' 16.9",     c=85.3596. 

9.  (7=46°  13' 20.9",    a=.0759588 
10.   (7=134°  36' 27.4",  6=27335.0. 


ANSWERS. 


123 


§  116;  page  70. 

3.  ^  =  28°  57' 18.0",  B=   46°  34'   2.8",  C=  104°  28 '39.0". 

4.  A  =  44°  24'  54.8",  B=   78°  27' 47.0",  C=    57°   7' 17.6". 

5.  A  =  71°  47'  24.4",  B=   58°  45'   5.4",  (7=   49°  27' 30.0". 

6.  A  =  74°  40 '16.4",  JS=   47°  46' 39.0",  C=    57°  33'   4.8". 

7.  A  =  59°  19' 11.8",  5=   68°  34'   7.6",  (7=   52°   6' 40.6". 

8.  ^  =  45°  11' 46.6",  B  =  101° 22'  17.8",  0=   33°25'56.4". 

9.  A  =  71°  33'  49.2".  10.   B=  30°  47'  22.8".     11.    O  =  25°  56 '54.2". 

§  121 ;  pages  73,  74. 

1.  B=   32°  36'   9.4",    c  =  6.62085. 

2.  B,=    31°  67' 47.8 ",  0!  =  120.313; 
£2  =  148°   2' 12.2",  «2  =  11.3800. 

3.  (7=    23°  33' 18.2",   a  =  .183882. 

4.  A  =   34°  29' 48.2",    6  =  7.12905. 

5.  Impossible. 

6.  Impossible. 

7.  B  =  48°  34'  38.4",  a  =  76.0172. 

8.  O=90°,  6  =  5.51109. 

9.  d  =    46°  18' 36.5",  (*!  =  6.94676 ; 
C2  =  133°  41'  24.5",  a2  =  .699906. 

10.  A=   25°  32' 50.9",    c  =  278.193. 

11.  Impossible. 

12.  O=  14° 4' 7.7",  6  =  1.43516. 

13.  5  =  90°,  c  =  137.872. 

14.  A,=   70°  12' 46.7",  6X  =  . 287904; 
A2  =  109°  47'  13.3",  62  =  .104539. 

15.  C=   45°  38' 30.2",   a  =  16214.3. 

§  122 ;  pages  74,  75. 

2.  197.656.          5.   165917.             8.   .078614.  11.  4000.81. 

3.  14.9812.          6.   2878.31.            9.   860.006.  12.  .000329015. 

4.  16.6843.          7.   1.30108.          10.   .0448746.  13.  25.6249. 

§  123 ;  pages  75,  76. 

1.  Height,  153.629  ft. ;  distances,  117.246  ft.,  217.246  ft. 

2.  AD  =  44.9525.  4.  47°  52' 2.1".  6.   56.6547,49.3482. 

3.  29799.9  sq.  rd.  5.   247.741  ft.  7.   35.2058  mi. 


124 


ANSWERS. 


8.  Two  angles,  74°  12' 20.0",  58°  23' 48.0";  third  side,  .430133. 

9.  K  47°  32'  33.1 "  W.  10.   9.8995  mi.,  19.1244  mi. 

11.   One  angle,  101°  13 '45.8";  diagonal,  136.187.         12.   297.954ft. 

13.  Sides,  26.5604,  90.5152 ;  one  angle,  119°  5f  14.6". 

14.  91.6364  ft.,  33.8973  ft.  15.   17.64934,8.77461. 
16.   1113.34ft.            17.   Diagonal,  52.9024;  side,  41.9505. 
18.  247.998ft.            19.   AD  =  88.1534,  ^1  =  56°  1'  10.7". 

20.   1569.948  sq.  rd. 

§  126 ;  page  79. 

2.  2.11491,  -1.86081,  -.254102.      4.   .47761,  -6.1364,  -.34120. 

3.  2.14510,  .523978,  -  2.66907.          5.   3.49086,  -  .83425,  .343379. 


5.  A 

6.  a 

7.  B 
or,  B 

8.  A 

9.  A 

10.  a 

11.  A 

or,  A 

12.  A: 

13.  A: 

14.  A: 

15.  A 

16.  a. 

17.  A 

18.  a 

19.  a 

20.  a 

21.  A 
or,  A 

22.  a 

23.  A 


:  36°  58'  50.0", 
;  27°  49' 17.9", 

68°  37' 18.1", 

111°  22' 41.9", 

;    68°  10'   4.4", 

15°  34' 32.3", 
170°  13' 25.6", 

21°  11  '12.7", 
158°  48' 47.3", 

82°   8' 19.3", 

122°  34' 33.5", 

153°  10'   2.8", 

165°  50' 26.0", 

112°  16' 49.7", 

:    55°  58'   5.5", 

:    54°   0'24.8", 

:   41°  29' 25.7", 

152°  35'  19.0", 

20°  3' 21.5", 
159°  56 '38.5", 
110°  57 '15.6", 
111°  53' 21.2", 


§  148 ;  page  93. 

B=    63°  42' 34.0", 

b=   42°  29' 21.8", 

6=   44°  56 '46.7", 

6  =  135°   3' 13.3", 

6  =  163°  42 '32.1", 

B=   94°  14' 40.0", 

B=    78°  34' 3.4", 

a=    19°  50' 30.4", 

a  =  160°   9' 29.6", 

a=    73°  38' 54.4", 

a  =  132°  24'  39.6", 

£  =  115°  25'   2.8", 

b  =  139°  10'  11.5", 

.  b  =  145°  51  '35.5", 

B=    34°  41 '20.4", 

B=    84°  43' 10.5", 

6  =  133°  39' 29  8", 

5  =  108°   7'   8.6", 

a=    14°  58 '21.1", 

a  =  165°   1'38.9", 

B=  165°  10'  31.9", 

JB  =  115°40'   6.8", 


6=  42°  34' 54.4". 
c=  49°  17' 42.4". 
c=  49°  20' 41.8"; 
c  =  130°  39 '18.2". 
c  =  141°  50' 15.2". 
c  =  105°  26' 27.5". 
b=  40°  1'  8.6". 
c=  69°  54' 41.6"; 
c  =  110°  5' 18.4". 
b=  28°  4' 23.5". 
B=  52°  58'  9.5". 
c=  20°  2' 40.3". 
c=  41°  42 '23.4". 
c=  71°  42 '41.1". 
c=  12°  39' 44.7". 
c=  86°  10' 32.3". 
c  =  121°  8 '21.5". 
b  =  125°  24'  13.7". 
c  =  131°  7'  4.9"; 
c=  48°  52' 55.1". 
c=  69°  41'  7.1". 
b  =  117°  49'  41.2". 


ANSWERS. 


125 


24.  .4  =  165°   3' 57.9", 

25.  B=   22°  13'   3.9", 
or,  B  =  157°  46'  56.1", 

26.  A=   64°  30'  52.0", 


2.  a  =  103°  25'  57.4", 

3.  a=  57°  43' 57.2", 

4.  .4  =  19°  56' 45.0", 

5.  A=  44°  41' 15.9", 

6.  B=  80°  27'  25.7", 
or,  5=  99°  32 '34.3", 

7.  A=  67°  11' 45.0", 


a  =  168°  8' 48.3", 
b=  20°  34' 38.3", 
b  =  159°  25'  21.7", 
a=  38°  32' 30.5", 

§  149  ;  page  94. 

J3  =  157°  31 '44.4", 
6  =  129°  56 '31,7", 

B  =  141°  38  '20.3", 
a=  51°  37'  1.9", 
b=  80°  46 '54.3", 
b=  99°  13'  5.7", 

B  =   80°  58' 16.5", 


6=    51°  53' 53.3". 

c  =  111°  38 '31.1"; 

c=    68°  21' 28.9". 

5  =146°  37' 27.3". 


C  =119°  19' 11.3". 
C=  58°  4' 55.6". 
6  =  113°  18' 58.3". 
1=  60*51'  3.4". 
C=  87°  31' 12.5"; 
C=  92°  28' 47.5". 
C  =  93°  29' 13.4". 


§150;  page  95. 

2.  o=    69°  55' 43.2",  C=  l59°59'40.6". 

3.  ^  =  120°  41' 19.6",    c=    30°  14' 37.4". 

4.  .4  =  140°  35'   4.5",  (7=  145°  11 '50.4". 

5.  C=  148°  19 '24.8",    c=    80°  47' 39.8". 


2.  a  =   95°  37' 51.0", 

3.  &=    98°  30' 32.4", 

4.  c=    64°  19 '27.8", 

5.  6  =  146°  25'   1.4", 


2.  A  =  121°  32'  41.3", 

3.  4=   86°  59' 48.8", 

4.  C=  134°  57' 31.3", 

5.  £  =  163°   8' 48.4", 


2.  .4=   51°  58' 28.0", 

3.  ^  =  142°  32 '37.8", 

4.  ^  =  142°  23' 44.0", 

5.  A=   47°  21' 11.8". 


§  161 ;  page  104. 

6=  41°  52 '22.2", 

c=  56°  42' 47.0", 

a=  34°   3' 11.8", 

a=  69°   4' 38.2", 

§  162  ;  page  105. 

B=  40°  56' 48.5", 
(7=  60°  50' 54.8", 
J5=  50°  40' 48.3", 
.4  =  147°  29'  24.2", 

§  163  ;  page  106. 

B=  58°  53' 13.2", 
B=  27°  52' 36.0", 
B  =  159°  15'  41.6", 


C=110°  48' 24.0". 
A=  59°  38 '53.2". 
B=  37°  39' 27.2". 
C=  125°  11  '41.8". 


c=  37°  25' 48.8". 
b  =  111°  16'  42.4". 
a=  69°  7' 34.6". 
c=  76°  8' 49.0". 


(7=  83°  54' 31.6". 
(7=  32°  26' 52.8". 
C=133°14'  4.2". 


126 


3.  a  =    68°  46' 28.4", 

4.  a=    90°  53'   2.6", 

5.  a  =  103°  31' 33.8", 

6.  6=   85°  48' 53.8". 

4.  C  =    65°  29'   1.0", 

5.  B=   42°  40'   9.2", 
or,  JB  =  137°  19'  50.8", 

6.  Impossible. 

7.  C=   90°, 

8.  £=    68°  17'   2.4", 
or,  B  =  111°  42'  57.6", 

9.  Impossible. 

10.  C  =146°  37'  40.2", 


ANSWERS. 

§  164  ;  page  107. 

b=  73°  47 '57.8", 
b  =  117°  48'  59.6", 
6=  53°  4' 26.2", 

§  165  ;  page  109. 
A=   97°  18' 33.8", 
(7=159°  54'   3.6", 
C  =    50°  21' 16.4", 

B  =  113°  33'  15.5", 
A  =  132°  35'  12.4", 
-4  =  77°  3' 48.0", 

B=   55°   l'H.8", 


§  166 ;  page  111. 

o=  82°  54'  0.0", 
c  =  160°  6' 10.0", 
c  =  103°  6' 20.4", 

B=   63°  46' 30.2", 


a  =  117°  9'  5.2", 
6  =  129°  9' 46.0", 
6=  20°  34' 54.2", 


c=  63°  12' 24.6". 
c  =  132°  5' 10.0". 
c  =  61°  14' 18.2". 


a  =  100°  42'  23.4". 

c  =  153°  29' 39.8"; 
c  =  90°  8' 51.4". 

6  =  114°  47'  47.5". 
a  =  131°  16'  32.2"; 
a=  95°  48' 41.8". 

b=   96°  33' 16.2". 


A=  79°  18' 29.0". 
C=164°  6'  8.4"; 
O=  128°  22' 54.8". 
b=  66°  29' 37.6". 


A=  47°  20' 57.2". 
B=  89°  23' 51.8"; 
B=  26°  57' 36.4". 


2.  b  =  114°  48'  57.9", 

3.  a=    67°  25'   2.3", 
or,  a  =  112°  34'  57.7", 

4.  c=    90°, 

5.  Impossible. 

6.  6  =   27°  22'   7.6", 

7.  a=   43°   2' 23.6", 
or,  a  =  136°  57' 36.4", 

8.  Impossible. 

§  167 ;  page  112. 

1.  Distance,  3275.20  mi. ;  bearing  of  Boston  from  Greenwich,  K  71 
38'  53.7"  W. ;  of  Greenwich  from  Boston,  K  53°  6'  31.9"  E. 

2.  Distance,  11012.9  mi. ;  bearing  of  Calcutta  from  Valparaiso,  S.  64 
20'  17.4"  E. ;  of  Valparaiso  from  Calcutta,  S.  54°  54'  25.2"  W. 

3.  Latitude,  49°  58'  23.1"  N. 

§  170 ;  page  113. 

1.  Time,  6  h.  0  m.  43  s.  A.M.  ;  longitude,  44°  49'  18"  W. 

2.  15°  0'  41.4".          3.   N.  56°  28'  8.5"  E.          4.   5  h.  3  m.  27  s.  A.M. 


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